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Calculus 1 : How to find differential functions

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Example Questions

Example Question #811 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(\pi x^4)$ at the point x=2

Possible Answers:

$-8\pi$

$8\pi$

$32\pi$

$-32\pi$

Correct answer:

$32\pi$

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function

$f(x)=sin(\pi x^4)$

The slope of the tangent is

$f'(x)=4\pi x^3 cos(\pi x^4)$

$f'(x)=4\pi (2)^3 cos(\pi (2)^4)=32\pi$

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Example Question #812 : Other Differential Functions

Determine the slope of the line that is tangent to the function f(x)=8x^2-5x+23 at the point x=3

Possible Answers:

Correct answer:

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

Taking the derivative of the function

f(x)=8x^2-5x+23

The slope of the tangent is

f'(x)=16x-5

f'(3)=16(3)-5=43

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Example Question #812 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function f(x)=9sin^3(x) at the point $x=\frac{\pi}{3}$

Possible Answers:

$\frac{27}{8}$

$-\frac{81}{8}$

$\frac{81}{8}$

$\frac{27\sqrt{3}}{8}$

$-\frac{27}{8}$

Correct answer:

$\frac{81}{8}$

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[sin(u)]=cos(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function

f(x)=9sin^3(x)

The slope of the tangent is

f'(x)=27cos(x)sin^2(x)

$f'(\frac{\pi}{3})=27cos(\frac{\pi}{3})sin^2(\frac{\pi}{3})=\frac{81}{8}$

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Example Question #1001 : Differential Functions

Determine the slope of the line that is tangent to the function f(x)=9cos^3(x) at the point $x=\frac{\pi}{3}$

Possible Answers:

$-\frac{9\sqrt{3}}{8}$

$-\frac{27\sqrt{3}}{8}$

$\frac{9\sqrt{3}}{8}$

$\frac{27}{8}$

$\frac{27\sqrt{3}}{8}$

Correct answer:

$-\frac{27\sqrt{3}}{8}$

Explanation:

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Trigonometric derivative:

d[cos(u)]=-sin(u)du

Note that u may represent large functions, and not just individual variables!

Taking the derivative of the function

f(x)=9cos^3(x)

The slope of the tangent is

f'(x)=-27sin(x)cos^2(x)

$f'(\frac{\pi}{3})=-27sin(\frac{\pi}{3})cos^2(\frac{\pi}{3})=\frac{-27\sqrt{3}}{8}$

Report an Error

Example Question #1002 : Differential Functions

Find the first derivative of $y=3x^2\cdot sin(x^2+1)$

Possible Answers:

y'=6x(x^2cos(x^2+1)+sin(x^2+1))

y'=-6xcos(x^2+1)

y'=6xcos(x^2+1)

y'=12x^2cos(x^2+1)

y'=3x(xcos(x^2+1)+2sin(x^2+1))

Correct answer:

y'=6x(x^2cos(x^2+1)+sin(x^2+1))

Explanation:

We must use the product rule here, which says

$\frac{d}{dx}\left [f(x)\cdot g(x) \right ]=f(x)g'(x)+f'(x)g(x)$

Here, f(x)=3x^2, g(x)=sin(x^2+1)

So, $y'=3x^2\cdot \left [ sin(x^2+1)\right ]'+(3x^2)'\cdotsin(x^2+1)$

Now, as we differentiate each term, we see that we will need the chain rule for the derivative in the first term, which says

$\frac{d}{dx}\left [f(g(x)) \right ]=f'(g(x))\cdot g'(x)$

Applying the rule, as we continue to differentiate

$y'=3x^2\cdot \left [ cos(x^2+1)\cdot (x^2+1)'\right ]+6x\cdot sin(x^2+1)$

$y'=3x^2\cdot \left [ cos(x^2+1)\cdot 2x\right ]+6x\cdot sin(x^2+1)$

Simplifying, we get

$y'=6x^3\cdot cos(x^2+1)+6x\cdot sin(x^2+1)$

y'=6x(x^2cos(x^2+1)+sin(x^2+1))

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Example Question #1003 : Differential Functions

Find the first derivative of y=(3x+1)^3

Possible Answers:

y'=9(3x+1)^2

y'=3(3x+1)^2

y'=3(3x+1)^4

y'=(3x+1)^2

y'=9(3x+1)^3

Correct answer:

y'=9(3x+1)^2

Explanation:

Here, we will need the chain rule, which says

$\frac{d}{dx}\left [f(g(x)) \right ]=f'(g(x))\cdot g'(x)$

Here, f(x)=x^3, g(x)=3x+1

Differentiating, we get

$y'=3(3x+1)^2\cdot 3$

y'=9(3x+1)^2

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Example Question #817 : Other Differential Functions

Find the first derivative of y=sin^2(x^2)

Possible Answers:

y'=4xsin(x^2)cos(x^2)

y'=2cos^2(x^2)sin(x^2)

y'=4xcos^2(x^2)

y'=2sin^2(x^2)

y'=4xsin^2(x^2)

Correct answer:

y'=4xsin(x^2)cos(x^2)

Explanation:

We need to use the chain rule TWICE, which says

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

Here,

f(x)=x^2, g(x)=sin(x), h(x)=x^2

Differentiating, gives us

$y'=2\cdot sin(x^2)\cdot cos(x^2)\cdot 2x$

Simplifying,

y'=4xsin(x^2)cos(x^2)

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Example Question #1004 : Differential Functions

Find the first derivative of y=x^2e^x

Possible Answers:

y'=2e^x

y'=2x+e^x

y'=xe^x(x+2)

y'=2xe^x+e^x

y'=2xe^x

Correct answer:

y'=xe^x(x+2)

Explanation:

We need to use the product rule, which says

$\frac{d}{dx}\left [f(x)\cdot g(x) \right ]=f(x)g'(x)+f'(x)g(x)$

Differentiating gives us

y'=x^2(e^x)'+(x^2)'e^x

y'=x^2e^x+2xe^x

Factoring, gives us

y'=xe^x(x+2)

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Example Question #1005 : Differential Functions

Find the first derivative of $y=3x\cdot tan(3x)$

Possible Answers:

y'=27xsec^2(3x)tan(3x)

y'=3sec^2(3x)

y'=9xsec^2(3x)tan(3x)

y'=9xsec^2(3x)+3tan(3x)

y'=3xsec^2(3x)

Correct answer:

y'=9xsec^2(3x)+3tan(3x)

Explanation:

We need the product rule, which says

$\frac{d}{dx}\left [f(x)\cdot g(x) \right ]=f(x)g'(x)+f'(x)g(x)$

Differentiating gives

$y'=3(x\cdot (tan(3x))'+(x)'\cdot tan(3x))$

To differentiate the second term, we need the chain rule, which says

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

Continuing to differentiate,

$y'=3(x\cdot sec^2(3x)\cdot 3+1\cdot tan(3x))$

Simplifying,

$y'=3(3x\cdot sec^2(3x)+tan(3x))$

y'=9xsec^2(3x)+3tan(3x)

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Example Question #820 : Other Differential Functions

Find the first derivative of $y=\frac{x^2-3}{x+4}$

Possible Answers:

y'=x^2+8x+3

y'=2x

$y'=\frac{2x}{x+4}$

$y'=\frac{x^2+8x+3}{(x+4)^2}$

$y'=\frac{x^2+8x+3}{x+4}$

Correct answer:

$y'=\frac{x^2+8x+3}{(x+4)^2}$

Explanation:

We need to use the quotient rule here to differentiate, which says

$\frac{d}{dx}\left [ \frac{f(x)}{g(x)} \right ]=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}$

Applying the quotient rule to differentiate,

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

$y'=\frac{(x+4)\cdot 2x-(x^2-3)\cdot 1}{(x+4)^2}$

Simplifying,

$y'=\frac{2x^2+8x-x^2+3}{(x+4)^2}$

$y'=\frac{x^2+8x+3}{(x+4)^2}$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #811 : How To Find Differential Functions

Example Question #812 : Other Differential Functions

Example Question #812 : How To Find Differential Functions

Example Question #1001 : Differential Functions

Example Question #1002 : Differential Functions

Example Question #1003 : Differential Functions

Example Question #817 : Other Differential Functions

Example Question #1004 : Differential Functions

Example Question #1005 : Differential Functions

Example Question #820 : Other Differential Functions

All Calculus 1 Resources

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