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Calculus 1 : Writing Equations

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Example Question #1292 : Functions

Find the first derivative of the following function.

f(x)=sin^2(x^3)

Possible Answers:

f'(x)=3x^2sin(x^3)cos(x^3)

f'(x)=6x^2sin(x^3)

f'(x)=6x^2cos(x^3)

None of the other answers.

f'(x)=6x^2sin(x^3)cos(x^3)

Correct answer:

f'(x)=6x^2sin(x^3)cos(x^3)

Explanation:

To find the derivative of a function in this form, we must make use of the chain rule, which states that for a give function:

F(x)=f(g(x))

Its derivative is defined as:

F'(x)=f'(g(x))*g'(x)

In this case we have the function:

F(x)=sin^2(x^3)=[sin(x^3)]^2

which is in the form of the composite function

F(x)=f(g(h(x)))

in which

f(x)=x^2 , g(x)=sinx , h(x)=x^3

By similar logic, the derivative of the composite function is

F'(x)=f'(g(h(x)))*g'(h(x))*h'(x)

in which

f'(g(h(x)))=2sin(x^3)

g'(h(x))=cos(x^3)

and h'(x)=3x^2

As a result, we have that

F'(x)=2sin(x^3)*cos(x^3)*3x^2=6x^2sin(x^3)cos(x^3)

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Example Question #1293 : Functions

Find the first derivative of the following function using the Product Rule.

f(x)=3x^2(2x^3+10x-11)

Possible Answers:

f'(x)=30x(x^3+2x-3)

f'(x)=18x^4+30x^2-6x

f'(x)=6(5x^4+15x^2-11x)

f'(x)=6x^5+30x^3-33x^2

None of the other answers are correct.

Correct answer:

f'(x)=6(5x^4+15x^2-11x)

Explanation:

The Product Rule of derivatives states that for a given function:

f(x)=uv

The derivative is defined as

f'(x)=u'v+v'u

In this case, for the given function

f(x)=3x^2(2x^3+10x-11)

u=3x^2 , v=2x^3+10x-11

and the respective derivatives are:

u'=6x , v'=6x^2+10

Applying the product rule we get that

f'(x)=(6x)*(2x^3+10x-11)+(6x^2+10)*(3x^2)

=(12x^4+60x^2-66x)+(18x^4+30x^2)

=30x^4+90x^2-66x=6(5x^4+15x^2-11x)

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Example Question #1294 : Functions

Find the first derivative of the given function.

$f(x)=\frac{x^2}{2x+3}$

Possible Answers:

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

None of the other answers

$f'(x)=\frac{3x^2-6x}{(2x+3)^2}$

$f'(x)=\frac{2x^2+5x}{(2x+3)^2}$

$f'(x)=\frac{3x^2+6x}{(2x+3)^3}$

Correct answer:

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

Explanation:

To find the first derivative of this function, we must make use of the Quotient Rule of derivatives. That is, for a function

$f(x)=\frac{u}{v}$

The derivative is defined as

$f'(x)=\frac{u'v-uv'}{v^2}$

In this case

u=x^2 , v=2x+3

u'=2x , v'=2

As a result,

$f'(x)=\frac{(2x)(2x+3)-(2)(x^2)}{(2x+3)^2}=\frac{4x^2+6x-2x^2}{(2x+3)^2}$

$=\frac{2x^2+6x}{(2x+3)^2}=\frac{2(x^2+3x)}{(2x+3)^2}$

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Example Question #1295 : Functions

Find the tangent line of the following function containing the point $\theta$ .

$y(\theta)=\cos(2\theta)+\sin(\theta)$

$\theta = -\frac{\pi}{2}$

Possible Answers:

The answer is not shown.

y=-2

$y=\frac{7}{2}x-2$

$y=\theta-2$

$y=2\theta$

Correct answer:

y=-2

Explanation:

First we note that

$f(-\frac{\pi}{2})=\cos(2(-\frac{\pi}{2}))+\sin(-\frac{\pi}{2})=\cos(\pi)-1=-1-1=-2$ .

Now we take the derivative of the function to find the slope of the tangent line.

Using the chain rule we get

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

$f'(\theta)=-2\sin{2\theta}+\cos{\theta}$ .

Now we want to find the specific slope of the tangent line containing the $\theta=-\frac{\pi}{2}$ .

We then find that the slope is

$f'(-\frac{\pi}{2})=-2\sin{(-2\frac{\pi}{2})}+\cos(\frac{-\pi}{2})=0+0=0$ .

We now plug in these values to slope-intercept form to find the y-intercept of the tangent line.

$-2=0\theta+b$

which gives us

b=-2 .

Thus the equation is simply

y=-2 .

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Calculus 1 : Writing Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1292 : Functions

Example Question #1293 : Functions

Example Question #1294 : Functions

Example Question #1295 : Functions

All Calculus 1 Resources

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