Differential Functions - AP Calculus AB

Card 0 of 14630

Question

Find the derivative at x=3.

Answer

First, find the derivative using the power rule:

$\frac{d}{dx}6x^3=18x^2$

Now, substitute 3 for x.

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Question

Find the derivative.

$\frac{2}{3x^4}$

Answer

Use the quotient rule to find the derivative.

$\frac{3x^4(0)-2(12x^3)}{(3x^4)^2}=\frac{-24x^3}{9x^8}$

Simplify.

The derivative is $\frac{-8}{3x^5}$ .

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Question

Find the derivative.

$\frac{1}{x^5}$

Answer

Use the quotient rule to find the derivative.

$\frac{x^5(0)-1(5x^4)}{(x^5)^2}=\frac{-5x^4}{x^{10}}=\frac{-5}{x^6}$

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Question

Find the derivative.

$12x^{12}$

Answer

Use the power rule to find the derivative.

$\frac{d}{dx}12x^{12}=144x^{11}$

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Question

Differentiate:

$f(x) = \left(\frac{1}{x}\right)\cos^2(x)$

Answer

To find the derivative of this function we must use the Product Rule and the Chain Rule. First we set

$h(x) = \frac{1}{x}$

and

$g(x) = \cos^2(x)$

Now differentiating both of these functions gives

$h'(x) = \frac{-1}{x^2}$

$g'(x) = 2\cos(x)(-\sin(x)) = -2\cos(x)\sin(x)$

Applying this to the Product Rule gives us,

$f'(x) = \frac{-1}{x^2}\cos^2(x)-\frac{2}{x}\cos(x)\sin(x)$

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Question

Differentiate the function

$f(x) = \sin(\cos(2x))$

Answer

To differentiate the function properly, we must use the Chain Rule which is,

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))g'(x)$

Therefore the derivative of the function is,

$f'(x) = \cos(\cos(2x))(-\sin(2x))(2) = -2\cos(\cos(2x))\sin(2x)$

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Question

Find the derivative.

$-\frac{5}{x^3}$

Answer

Use the quotient rule to find this derivative.

Recall the quotient rule:

$\ \ h(x)=\frac{f(x)}{g(x)} \ \ h'(x)=\frac{g(x)f'(x)-g'(x)f(x)}{(g(x))^2}$

$h'(x)=\frac{x^3(0)-(-5)3x^2}{x^6}=\frac{15x^2}{x^6}=\frac{15}{x^4}$

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Question

Differentiate

$f(x) = \sin^2(x^2)$

Answer

To differentiate this equation we use the Chain Rule.

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Using this throughout the equation gives us,

$f'(x) = 2\sin(x^2)\cos(x^2)(2x)$

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Question

Find the derivative of

$f(x) = 4\sin(x+3x^2)$

Answer

To find the derivative of the function we must use the Chain Rule, which is

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we get,

$f'(x) = 4\cos(x+3x^2)(1+6x)$

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Question

Find the slope of the function at .

Answer

To consider finding the slope, let's discuss the topic of the gradient.

For a function , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rules will be necessary:

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

at

x:

$\frac{\delta f}{\delta x}=12x^2+6xy+2y^2;12(1)^2+6(1)(1)+2(1)^2=20$

y:

$\frac{\delta f}{\delta y}=3x^2+4xy+3y^3;3(1)^2+4(1)(1)+3(1)^3=10$

The slope is

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Question

Find the derivative of

$f(x) =8\sin(4x+3)$

Answer

To find the derivative of the function we must use the Chain Rule

$f(x) = h(g(x)) \rightarrow f'(x) = h'(g(x))\cdot g'(x)$

Applying this to the function we are given gives,

$f'(x) = 8\cos(4x+3)(4) = 32\cos(4x+3)$

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Question

Using the method of midpoint Reimann sums, approximate the area of the region between the functions $f(x)=10(1-e^{-x})$ and over the interval using three midpoints.

Answer

The Reimann sum approximation of an integral of a function with subintervals over an interval takes the form:

$\int_a^b f(x)dx \approx \frac{b-a}{n}[f(x_1)+f(x_2)+...+f(x_n)]$

Where $\frac{b-a}{n}$ is the length of the subintervals.

For this problem, since there are three midpoints, the subintervals have length $\frac{3-0}{3}=1$ , and the midpoints are .

Furthermore, since the question is asking for the area between the two functions, it's asking for the difference between the larger function over the interval, f(x), and the smaller function, g(x).

The integral is thus:

${\int_0^{3} (f(x)-g(x))dx \approx [(10(1-e^{-0.5})-2(0.5))+(10(1-e^{-1.5})-2(1.5))+(10(1-e^{-2.5})-2(2.5))]}$

${\int_0^{3} (f(x)-g(x))dx \approx 11.8825$

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Question

A Riemann Sum approximation of an integral $\int_a^b (f(x))dx$ follows the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$ .

Where n is number of points/subintervals used to approximate the integral.

Knowing this, imagine a modified style of Riemann Sum, such that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$ .

Which of the following parameters would give the closest integral approximation of the function:

$\int_{10}^{100} \frac{e^x}{x^2}dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=20 will give a closer approximation than n=10:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

Consider the function $\frac{e^x}{x^2}$ over the interval . The steepness of this graph at a point can be found by taking the function's derivative.

The quotient rule of derivatives states:

$d\frac{u}{v}=\frac{vdu-udv}{v^2}$

So the derivative is

$\frac{x^2e^x-2xe^x}{x^4}$

$\frac{e^x(x-2)}{x^3}$

It might be intuitive to see that the steepness is positive and that it gets progressively greater over the specified interval, although to be precise, we may take the derivative once more to find the rate of change of this steepness and see if it's postive or negative:

$\frac{x^4(2xe^x+x^2e^x-2e^x-2xe^x)-4x^3(x^2e^x-2xe^x)}{x^8}$

$\frac{x^6e^x+6x^4e^x-4x^5e^x}{x^8}$

$\frac{e^x(x^2-4x+6)}{x^4}$

Since this derivative is positive over the specified interval, , and since the slope was shown to be positive as well, the slope is ever-increasing and growing steeper.

The intervals should in turn grow increasingly thinner.

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Question

The general Reimann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_0^2 (x^3-6x^2+12x+6)dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so n=50 will give a closer approximation than n=40:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

First find the derivative of the function :

This can be rewritten as

Thus for all values for x outside of 2, the slope is positive, though as the function is continuous, a single point is nigh negligible.

Now take the second derivative:

For the interval , the second derivative is negative.

Since the signs of the first and second derivatives are opposing, the function is flattening out. The intervals should grow wider.

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Question

The general Reimann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{0.1}^{0.01} x\ln(x)dx$ ?

Answer

The more points/intervals that are selected, the closer the Reimann sum approximation becomes to the actual integral value, so will give a closer approximation than :

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

For the function

$x\ln(x)$

The derivative can be found, using the product rule :

$\ln(x)+x\frac{1}{x}$

$\ln(x)+1$

Over the entire interval , this function is negative. Now, consider the second derivative:

$\frac{1}{x}$

Over the interval , the function is positive.

Since the first and second derivatives have opposing signs, the function is flattening out. The subintervals should grow progressively thicker.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_0^{0.5} \frac{1}{2}^{x^2-2x+1}dx$ ?

Answer

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=\frac{1}{2}^{x^2-2x+1}$ , find the first derivative:

$f'(x)=(2x-2)\frac{1}{2}^{x^2-2x+1}ln(\frac{1}{2})$

The term in the parenthesis is what may lend to variability in the sign. Over the interval the parenthetical is negative, so the derivative is positive, since $ln(\frac{1}{2})=-0.693$ , and the product of two negatives is positive. Now consider the second derivative:

$f''(x)=\frac{1}{2}^{x^2-2x+2}ln(\frac{1}{2})+(2x-2)^2\frac{1}{2}^{x^2-2x+1}ln^2(\frac{1}{2})$

The second term will always be positive, and the first term will always be negative. On the interval , it can be found, however, that the second derivative will be positive. If calculating this is to nettlesome, plot this function on the interval:

Since the first and second derivatives have opposite signs, the function is getting flatter. The intervals should in turn get wider.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_0^{\frac{\pi}{2}} cos(\frac{x^2}{\pi})dx$ ?

Answer

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)=cos(\frac{x^2}{\pi})$ take the first derivative:

$f'(x)=-\frac{2x}{\pi}sin(\frac{x^2}{\pi})$

Within the interval $x=[0,\frac{\pi}{2}]$ this function is negative. Now consider the second derivative:

$f''(x)=-\frac{2}{\pi}sin(\frac{x^2}{\pi})-\frac{4x^2}{\pi^2}cos(\frac{x^2}{\pi})$

Over the interval $x=[0,\frac{\pi}{2}]$ , the second derivative is negative as well.

Since the signs match, the slope gets steeper, and the intervals should get thinner.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{\frac{\pi}{3}}^{\frac{\pi}{2}} e^{cos(x)}dx$ ?

Answer

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering $f(x)= e^{cos(x)}$ , find the first derivative:

$f'(x)=-sin(x)e^{cos(x)}$

Within the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , the function only takes negative values. Now, take the second derivative:

$f''(x)=-cos(x)e^{cos(x)}+sin^2(x)e^{cos(x)}$

On the interval $x=[\frac{\pi}{3},\frac{\pi}{2}]$ , this function takes on positive values.

Since the signs of the derivatives are opposing, the function gets flatter over the course of this interval, and the subintervals should thus get wider.

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Question

The general Riemann Sum approximation of an integral $\int_a^b f(x)dx$ takes the form

$\frac{b-a}{n}\sum_{i=1}^{n}f(x_i)$

Where is the number of points/subintervals, and each subinterval is of uniform width $\frac{b-a}{n}$ .

Knowing this, imagine that the subintervals are not of uniform width.

Denoting a particular subinterval's width as ,

the integral approximation becomes

$\sum_{i=1}^{n}I_if(x_i)$

Which of the following parameters would give the closest integral approximation of the function:

$\int_{ln(\frac{\pi}{2})}^{ln(\pi)} sin(e^{x})dx$ ?

Answer

The more points/intervals that are selected, the closer the Riemann sum approximation becomes to the actual integral value, so n=16 will give a closer approximation than n=8:

Now, assuming the amount of points/intervals are fixed, but the actual widths can differ, it would be prudent to have narrower intervals at those regions where the graph is steepest. Relatively flatter regions of the function do not lead to the approximation overshooting or undershooting as much as the steeper areas, as can be seen in the above figure.

To determine how the widths of the intervals should change if they should, determine whether or not the function is getting steeper or flatter. The function will be getting steeper over an interval if the 1st and 2nd derivatives have the same sign over this interval.

The first derivative is the slope. The second derivative is the rate of change of the slope. If the rate of change has the same sign as the slope, it is pushing the absolute value of the slope away from 0. Since 0 indicates flatness, this means that the function is growing steeper. The opposite holds if the signs are opposite!

Considering the function $f(x)=sin(e^{x})$ , take the first derivative:

$f'(x)=e^xcos(e^{x})$

Over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ this function is negative due to the cosine term. Now take the second derivative:

$f''(x)=e^xcos(e^{x})-e^{2x}sin(x)$

This function is also negative over the interval $x=[ln(\frac{\pi}{2}),ln(\pi)]$ .

Since both derivatives share a sign, the function grows steeper over the specified interval. The subintervals should therefore get progressively narrower.

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Question

Differentiate the function:

Answer

To differentiate this problem we will need to use the power rule.

The power rule is, where n is the exponent.

Thus our derivative is,

$f'(z)=\frac{d}{dz}(az^-^7)=-7\cdot az^-^7^-^1=-7az^-^8$ .

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