AP Calculus BC - Derivatives

What is the equation of the line tangent to the graph of the function

at ?

Explanation

The slope of the line tangent to the graph of at is

, which can be evaluated as follows:

$f'(x)= 3x^{3-1}-0$

$f'(x)= 3x^{2}$

Then $f'(2)= 3 \cdot 2^{2} = 3 \cdot4 = 12$ , which is the slope of the line.

The equation of the line with slope 12 through is:

$y - y_{1} = m (x-x_{1})$

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\&y=130ln(9x)e^{(-12x)}\end{align*}$

$\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}$

$-\frac{(12e^{(-12x)})}{x}$

$\frac{(120e^{(-12x)})}{x}+ 1404ln(9x)e^{(-12x)}$

$\frac{1}{x}- 12e^{(-12x)}$

Explanation

$\begin{align*}&\text{The function that we have been given can be seen as two smaller functions multiplied together,}\&\text{the first being: }e^{(-12x)}\&\text{and the second being: }ln(9x)\&\text{With the product scaled by }130\&\text{To perform the derivative, }\text{ we'll wish to use the product rule, }d[uv]=udv+vdu\&\text{along with the following:}\&d[e^{ax}]=ae^{ax}dx\&d[ln(ax)]=\frac{dx}{x}\&u=e^{(-12x)};du=-12e^{(-12x)}\&v=ln(9x);dv=\frac{1}{x}\&\text{So the derivative is:}\&\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}\end{align*}$

Find the derivative of $\small lnx^{lnx}$ .

$\small \small lnx^{lnx}(ln(lnx)/x+1/x)$

$\small \small lnx^{lnx}ln(lnx)/x$

$\small lnxlnx/x$

$\small (lnxlnx^{lnx-1})/x$

$\small \small lnx^{lnx}(ln(lnx)+1/x)$

Explanation

To solve this derivative, we need to use logarithmic differentiation. This allows us to use the logarithm rule $\small lna^{b}=blna$ to solve an easier derivative.

Let $\small y=lnx^{lnx}$ .

Now we'll take the natural log of both sides to get

$\small lny=ln(lnx^{lnx})=lnxln(lnx)$ .

Now we can use implicit differentiation to solve for $\small y'$ .

The derivative of $\small lny$ is $\small y'/y$ , and the derivative of $\small \small \small lnx(ln(lnx))$ can be found using the product rule, which states

$\small \frac{\mathrm{d} }{\mathrm{d} x}(u\cdot v)=u'v+uv'$ where $\small u$ and $\small v$ are functions of $\small x$ .

Letting $\small u=lnx$ and $\small v=ln(lnx)$

(which means $\small u'=1/x$ and $\small v'=1/(xlnx)$ ) we get our derivative to be $\small ln(lnx)/x+1/x$ .

Now we have $\small y'/y=ln(lnx)/x+1/x$ , but $\small y=lnx^{lnx}$ , so subbing that in we get

$\small \small y'/(lnx^{lnx})=ln(lnx)/x+1/x$ .

Multiplying both sides by $\small lnx^{lnx}$ , we get

$\small y'=lnx^{lnx}(ln(lnx)/x+1/x)$ .

That is our derivative.

If , what is $\frac{d^2 u}{dx^2}$ ?

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

Explanation

To find the second derivative, first one has to find the first derivative, then take the derivative of this result.

The derivative of is .

The derivative of is , and this is our final answer.

Find the second derivative of the following equation:

$f''(x)=-\frac{4}{(2x-1)^2}$

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

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

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

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

Explanation

To find the second derivative, first we need to find the first derivative. The derivative of a natural log is the derivative of operand times the inverse of the operand. So for the given function, we get the first derivative to be

$f'(x)=2\cdot \frac{1}{2x-1}=\frac{2}{2x-1}$ .

Now, we have to take the derivative of the first derivative. To simplify this, we can rewrite the function to be . From here we can use the chain rule to solve for the derivative. First, multiply by the exponent and find the new exponent by subtracting the old one by one. Next multiply by the derivative of (2x-1) and then simplify. Thus, we get

$f''(x)=2\cdot-1\cdot2(2x-1)^-^2=-\frac{4}{(2x-1)^2}$ .

Find the derivative of the following function at :

$f(x)=\frac{e^x\cos(x)}{3}$

$\frac{1}{3}$

$-\frac{1}{3}$

Explanation

The derivative of the function is

$f'(x)=\frac{1}{3}(e^x\cos(x)-e^x\sin(x))$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ ,

$\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

Evaluated at the point x=0, we get

$f'(0)=\frac{1}{3}(1-0)=\frac{1}{3}$ .

What is the slope of $f(x)=x^{2}-3x+7$ at the point ?

Explanation

We define slope as the first derivative of a given function.

Since we have $f(x)=x^{2}-3x+7$ , we can use the Power Rule

$\frac{d}{dx}x^{n}=nx^{n-1}$ for all to determine that

$f'(x)=(2)x^{2-1}-(1)3x^{1-1}+(0)7=2x-3$ .

We also have a point with a -coordinate , so the slope

Use a definition of the derivative with the function to evaluate the following limit:

$\lim_{h \to 0}\frac{a^h-1}{h}$

$\ln(a)$

Explanation

Using the definition $f'(x) = \lim_{h \to 0}\frac{f(x+h)-f(x)}{h}$

And plugging in our function, we get that

$(a^x)' = \lim_{h \to 0}\frac{a^{x+h}-a^x}{h}$ .

if we factor out inside the limit we get $\lim_{h\to 0} a^x(\frac{a^h-1}{h})$

since the term doesn't contain an h we can factor it out, and then divide by both sides, getting that

$\frac{(a^x)'}{a^x} = \lim_{h\to 0}\frac{a^h-1}{h}$

but we know that $(a^x)' = a^x \cdot \ln(a)$

So we find that the limit is equal to $\ln(a)$ .

Using the limit definition of a derivative, find the derivative of the following function at :

Explanation

The limit definition of a derivative is

$\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$

where h is a very small change in x.

Using the above formula but with the function given, we get

$\lim_{h\rightarrow 0}\left[\frac{(x+h)+1-(x+1)}h\right]$

which simplified becomes

$\lim_{h\rightarrow 0}\frac{h}{h}=1$

Regardless of the x-value of the function, the derivative will always be 1 (the above contains no x).

$\begin{align*}&\text{Take the derivative with respect to }x\text{ of the function}\&y=130ln(9x)e^{(-12x)}\end{align*}$

$\frac{(130e^{(-12x)})}{x}- 1560ln(9x)e^{(-12x)}$

$-\frac{(12e^{(-12x)})}{x}$

$\frac{(120e^{(-12x)})}{x}+ 1404ln(9x)e^{(-12x)}$

$\frac{1}{x}- 12e^{(-12x)}$

Derivatives

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