Using the Chain Rule - Math

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Question

If what is the slope of the line at .

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Answer

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at. So, to find the derivative we will need to use the chain rule. The chain rule says

$$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x)$

so if we let and then

since $f'(x)=\frac{1}{x}$}$ and

$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x) = $$\frac{1}{x^3$$}$*3x^2$ = $\frac{3}{x}$

Therefore we evaluate at and we get $\frac{3}{12}$ or .

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Question

Find the derivative of the following function:

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Answer

Use -substitution so that .

Then the function becomes .

By the chain rule, $f'(x) = $\frac{du}{dx}$ \times $\frac{df}{du}$$ .

We calculate each term using the power rule:

$$\frac{du}{dx}$ = 2$

$$\frac{df}{du}$ = 2u$

Plug in :

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Question

What is the first derivative of ?

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Answer

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $$\frac{\mathrm{d}$ }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$

Plug in our equations.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$$=(2(x+6)^{1}$)(1+0)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=(2(x+6))$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=2x+12$

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Question

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)=?$

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Answer

For this problem we need to use the chain rule:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)$\frac{\mathrm{d}$ }{\mathrm{d} x}2x$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\cos(2x)2$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=2\cos(2x)$

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Question

If what is the slope of the line at .

Tap to reveal answer

Answer

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at. So, to find the derivative we will need to use the chain rule. The chain rule says

$$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x)$

so if we let and then

since $f'(x)=\frac{1}{x}$}$ and

$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x) = $$\frac{1}{x^3$$}$*3x^2$ = $\frac{3}{x}$

Therefore we evaluate at and we get $\frac{3}{12}$ or .

← Didn't Know|Knew It →

Question

Find the derivative of the following function:

Tap to reveal answer

Answer

Use -substitution so that .

Then the function becomes .

By the chain rule, $f'(x) = $\frac{du}{dx}$ \times $\frac{df}{du}$$ .

We calculate each term using the power rule:

$$\frac{du}{dx}$ = 2$

$$\frac{df}{du}$ = 2u$

Plug in :

← Didn't Know|Knew It →

Question

What is the first derivative of ?

Tap to reveal answer

Answer

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $$\frac{\mathrm{d}$ }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$

Plug in our equations.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$$=(2(x+6)^{1}$)(1+0)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=(2(x+6))$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=2x+12$

← Didn't Know|Knew It →

Question

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)=?$

Tap to reveal answer

Answer

For this problem we need to use the chain rule:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)$\frac{\mathrm{d}$ }{\mathrm{d} x}2x$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\cos(2x)2$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=2\cos(2x)$

← Didn't Know|Knew It →

Question

If what is the slope of the line at .

Tap to reveal answer

Answer

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at. So, to find the derivative we will need to use the chain rule. The chain rule says

$$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x)$

so if we let and then

since $f'(x)=\frac{1}{x}$}$ and

$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x) = $$\frac{1}{x^3$$}$*3x^2$ = $\frac{3}{x}$

Therefore we evaluate at and we get $\frac{3}{12}$ or .

← Didn't Know|Knew It →

Question

Find the derivative of the following function:

Tap to reveal answer

Answer

Use -substitution so that .

Then the function becomes .

By the chain rule, $f'(x) = $\frac{du}{dx}$ \times $\frac{df}{du}$$ .

We calculate each term using the power rule:

$$\frac{du}{dx}$ = 2$

$$\frac{df}{du}$ = 2u$

Plug in :

← Didn't Know|Knew It →

Question

What is the first derivative of ?

Tap to reveal answer

Answer

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $$\frac{\mathrm{d}$ }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$

Plug in our equations.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$$=(2(x+6)^{1}$)(1+0)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=(2(x+6))$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=2x+12$

← Didn't Know|Knew It →

Question

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)=?$

Tap to reveal answer

Answer

For this problem we need to use the chain rule:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)$\frac{\mathrm{d}$ }{\mathrm{d} x}2x$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\cos(2x)2$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=2\cos(2x)$

← Didn't Know|Knew It →

Question

If what is the slope of the line at .

Tap to reveal answer

Answer

The slope at any point on a line is also equal to the derivative. So first we want to find the derivative function of this function and then evaluate it at. So, to find the derivative we will need to use the chain rule. The chain rule says

$$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x)$

so if we let and then

since $f'(x)=\frac{1}{x}$}$ and

$\frac{d}{dx}$} f(g(x)) = f'(g(x))g'(x) = $$\frac{1}{x^3$$}$*3x^2$ = $\frac{3}{x}$

Therefore we evaluate at and we get $\frac{3}{12}$ or .

← Didn't Know|Knew It →

Question

Find the derivative of the following function:

Tap to reveal answer

Answer

Use -substitution so that .

Then the function becomes .

By the chain rule, $f'(x) = $\frac{du}{dx}$ \times $\frac{df}{du}$$ .

We calculate each term using the power rule:

$$\frac{du}{dx}$ = 2$

$$\frac{df}{du}$ = 2u$

Plug in :

← Didn't Know|Knew It →

Question

What is the first derivative of ?

Tap to reveal answer

Answer

To solve for the first derivative, we're going to use the chain rule. The chain rule says that when taking the derivative of a nested function, your answer is the derivative of the outside times the derivative of the inside.

Mathematically, it would look like this: $$\frac{\mathrm{d}$ }{\mathrm{d} x}f(g(x))=f'(g(x))g'(x)$

Plug in our equations.

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$$=(2(x+6)^{1}$)(1+0)$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=(2(x+6))$

$$\frac{\mathrm{d}$ }{\mathrm{d} $x}(x+6)^2$=2x+12$

← Didn't Know|Knew It →

Question

$$\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)=?$

Tap to reveal answer

Answer

For this problem we need to use the chain rule:

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\frac{\mathrm{d}$ }{\mathrm{d} x} \sin(2x)$\frac{\mathrm{d}$ }{\mathrm{d} x}2x$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=\cos(2x)2$

$$\frac{\mathrm{d}$ }{\mathrm{d} x}\sin(2x)=2\cos(2x)$

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