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Calculus 1 : Differential Functions

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Example Questions

Example Question #832 : Other Differential Functions

Find the first derivative of the following function:

$f(x)=3xe^{x}+12\cos(x)$ $\displaystyle f(x)=3xe^{x}+12\cos(x)$

Possible Answers:

$3+3xe^x-12\sin(x)$ $\displaystyle 3+3xe^x-12\sin(x)$

$3e^x+3xe^x+12\sin(x)$ $\displaystyle 3e^x+3xe^x+12\sin(x)$

$3e^x+3xe^x-12\sin(x)$ $\displaystyle 3e^x+3xe^x-12\sin(x)$

$3e^x+3xe^x-\sin(x)$ $\displaystyle 3e^x+3xe^x-\sin(x)$

$3e^x+3x^2e^x-12\sin(x)$ $\displaystyle 3e^x+3x^2e^x-12\sin(x)$

Correct answer:

$3e^x+3xe^x-12\sin(x)$ $\displaystyle 3e^x+3xe^x-12\sin(x)$

Explanation:

The derivative of the function is equal to

$f'(x)=3e^x+3xe^x-12\sin(x)$ $\displaystyle f'(x)=3e^x+3xe^x-12\sin(x)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)$, $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$, $\frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \cos(x)=-\sin(x)$

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Example Question #832 : How To Find Differential Functions

Find the derivative of the following function at x=1:

$(f\circ g)(x)$ $\displaystyle (f\circ g)(x)$

where $\displaystyle f(x)=2x^2$, g(x)=5x $\displaystyle g(x)=5x$

Possible Answers:

100 $\displaystyle 100$

$\displaystyle 0$

$\displaystyle 20$

$\displaystyle 50$

Correct answer:

100 $\displaystyle 100$

Explanation:

To find the derivative of the composite function, we must rewrite the composition as f(g(x)) $\displaystyle f(g(x))$. Using the chain rule, the derivative of this function is equal to

$f'(g(x)) \cdot g'(x)$ $\displaystyle f'(g(x)) \cdot g'(x)$.

For our function,

$\displaystyle f'(x)=4x$ and g'(x)=5 $\displaystyle g'(x)=5$. These derivatives were found using the following rule:

$\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$.

We now have all the pieces to use the chain rule formula, where we are evaluating the function at x=1:

g(1)=5 $\displaystyle g(1)=5$

g'(1)=5 $\displaystyle g'(1)=5$

$\displaystyle f'(g(1))=f'(5)=20$

$20 \cdot 5=100$ $\displaystyle 20 \cdot 5=100$

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Example Question #834 : Other Differential Functions

Find the first derivative of the function:

$f(x)= \frac{t^4-t^2}{e^t}$ $\displaystyle f(x)= \frac{t^4-t^2}{e^t}$

Possible Answers:

$\displaystyle e^t(-t^4+4t^3+t^2-2t)$

$\displaystyle e^t(t^4+4t^3+t^2+2t)$

$\displaystyle e^t(2t-t^2)$

$\frac{-t^4+4t^3+t^2-2t}{e^{2t}}$ $\displaystyle \frac{-t^4+4t^3+t^2-2t}{e^{2t}}$

Correct answer:

$\displaystyle e^t(-t^4+4t^3+t^2-2t)$

Explanation:

The derivative of the function is equal to

$f'(x)=\frac{e^t(4t^3-2t)-e^t(t^4+t^2)}{e^{2t}}=e^t(-t^4+4t^3+t^2-2t)$ $\displaystyle f'(x)=\frac{e^t(4t^3-2t)-e^t(t^4+t^2)}{e^{2t}}=e^t(-t^4+4t^3+t^2-2t)$

and was found using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}$, $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$, $\frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^x=e^x$

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Example Question #1021 : Differential Functions

Find the second derivative of the function:

$f(x)=\cos(\sec(e^{10x^2}))$ $\displaystyle f(x)=\cos(\sec(e^{10x^2}))$

Possible Answers:

$\displaystyle 1$

$20xe^{10x^2}\sec(x)\tan(x)$ $\displaystyle 20xe^{10x^2}\sec(x)\tan(x)$

$20xe^{10x^2}$ $\displaystyle 20xe^{10x^2}$

$-20xe^{10x^2}$ $\displaystyle -20xe^{10x^2}$

$\displaystyle 0$

Correct answer:

$20xe^{10x^2}$ $\displaystyle 20xe^{10x^2}$

Explanation:

Because $\cos(x)$ $\displaystyle \cos(x)$ and $\sec(x)$ $\displaystyle \sec(x)$ are inverses of each other, we can simplify the given function to

$f(x)=e^{10x^2}$ $\displaystyle f(x)=e^{10x^2}$

Taking the derivative of this, we get

$f'(x)=20xe^{10x^2}$ $\displaystyle f'(x)=20xe^{10x^2}$

by using the following rules:

$\frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}$, $\frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}$

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Example Question #832 : Other Differential Functions

Determine the slope of the line that is tangent to the function $\displaystyle f(x)=sin(10x)$ at the point $x=\pi$ $\displaystyle x=\pi$

Possible Answers:

$\displaystyle 0$

$\displaystyle -1$

$\displaystyle -10$

$\displaystyle 10$

$\displaystyle 1$

Correct answer:

$\displaystyle -10$

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:

$\displaystyle d[sin(u)]=cos(u)du$

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

Taking the derivative of the function $\displaystyle f(x)=sin(10x)$ at the point $x=\pi$ $\displaystyle x=\pi$

The slope of the tangent is

$\displaystyle f'(x)=10cos(x)$

$f'(\pi)=10cos(\pi)=-10$ $\displaystyle f'(\pi)=10cos(\pi)=-10$

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Example Question #1021 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=sin(5\pi x^2)$ $\displaystyle f(x)=sin(5\pi x^2)$ at the point x=2 $\displaystyle x=2$

Possible Answers:

$\displaystyle 0$

$20\pi$ $\displaystyle 20\pi$

$-10\pi$ $\displaystyle -10\pi$

$-20\pi$ $\displaystyle -20\pi$

$10\pi$ $\displaystyle 10\pi$

Correct answer:

$20\pi$ $\displaystyle 20\pi$

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:

$\displaystyle d[sin(u)]=cos(u)du$

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

Taking the derivative of the function $f(x)=sin(5\pi x^2)$ $\displaystyle f(x)=sin(5\pi x^2)$ at the point x=2 $\displaystyle x=2$

The slope of the tangent is

$f'(x)=10\pi x cos(5\pi x^2)$ $\displaystyle f'(x)=10\pi x cos(5\pi x^2)$

$f'(2)=10\pi (2) cos(5\pi (2)^2)=20\pi$ $\displaystyle f'(2)=10\pi (2) cos(5\pi (2)^2)=20\pi$

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Example Question #837 : Other Differential Functions

Determine the slope of the line that is tangent to the function $\displaystyle f(x)=sin^2(5x)$ at the point $x=\pi$ $\displaystyle x=\pi$

Possible Answers:

$\displaystyle 10$

$\displaystyle 0$

$\displaystyle -5$

$\displaystyle -10$

$\displaystyle 5$

Correct answer:

$\displaystyle 0$

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:

$\displaystyle d[sin(u)]=cos(u)du$

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

Taking the derivative of the function $\displaystyle f(x)=sin^2(5x)$ at the point $x=\pi$ $\displaystyle x=\pi$

The slope of the tangent is

$\displaystyle f'(x)=10sin(5x)cos(5x)$

$f'(\pi)=10sin(5\pi)cos(5\pi)=0$ $\displaystyle f'(\pi)=10sin(5\pi)cos(5\pi)=0$

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Example Question #841 : Other Differential Functions

Determine the slope of the line that is tangent to the function $\displaystyle f(x)=cos(4x)$ at the point $x=\frac{\pi}{2}$ $\displaystyle x=\frac{\pi}{2}$

Possible Answers:

$\displaystyle -4$

$\displaystyle 4$

$\displaystyle 1$

$\displaystyle -1$

$\displaystyle 0$

Correct answer:

$\displaystyle 0$

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:

$\displaystyle d[cos(u)]=-sin(u)du$

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

Taking the derivative of the function $\displaystyle f(x)=cos(4x)$ at the point $x=\frac{\pi}{2}$ $\displaystyle x=\frac{\pi}{2}$

The slope of the tangent is

$\displaystyle f'(x)=-4sin(4x)$

$f'(\frac{\pi}{2})=-4sin(4(\frac{\pi}{2}))=0$ $\displaystyle f'(\frac{\pi}{2})=-4sin(4(\frac{\pi}{2}))=0$

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Example Question #1021 : Differential Functions

Determine the slope of the line that is tangent to the function $f(x)=cos^2(\pi x^2)$ $\displaystyle f(x)=cos^2(\pi x^2)$ at the point $x=\frac{1}{2}$ $\displaystyle x=\frac{1}{2}$

Possible Answers:

$\displaystyle 0$

$\pi$ $\displaystyle \pi$

$-4\pi$ $\displaystyle -4\pi$

$-\pi$ $\displaystyle -\pi$

$-2\pi$ $\displaystyle -2\pi$

Correct answer:

$-\pi$ $\displaystyle -\pi$

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:

$\displaystyle d[cos(u)]=-sin(u)du$

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

Taking the derivative of the function $f(x)=cos^2(\pi x^2)$ $\displaystyle f(x)=cos^2(\pi x^2)$ at the point $x=\frac{1}{2}$ $\displaystyle x=\frac{1}{2}$

The slope of the tangent is

$f'(x)=-4\pi x sin(\pi x^2)cos(\pi x^2)$ $\displaystyle f'(x)=-4\pi x sin(\pi x^2)cos(\pi x^2)$

$f'(\frac{1}{2})=-4\pi (\frac{1}{2}) sin(\pi (\frac{1}{2})^2)cos(\pi (\frac{1}{2})^2)=-\pi$ $\displaystyle f'(\frac{1}{2})=-4\pi (\frac{1}{2}) sin(\pi (\frac{1}{2})^2)cos(\pi (\frac{1}{2})^2)=-\pi$

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Example Question #841 : How To Find Differential Functions

Determine the slope of the line that is tangent to the function $\displaystyle f(x)=cos(sin(x))$ at the point $x=3\pi$ $\displaystyle x=3\pi$

Possible Answers:

$\displaystyle -1$

$\displaystyle 3$

$\displaystyle -3$

$\displaystyle 0$

$\displaystyle 1$

Correct answer:

$\displaystyle 0$

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:

$\displaystyle d[sin(u)]=cos(u)du$

$\displaystyle d[cos(u)]=-sin(u)du$

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

Taking the derivative of the function $\displaystyle f(x)=cos(sin(x))$ at the point $x=3\pi$ $\displaystyle x=3\pi$

The slope of the tangent is

$\displaystyle f'(x)=cos(x)sin(sin(x))$

$f'(3\pi)=cos(3\pi)sin(sin(3\pi))=0$ $\displaystyle f'(3\pi)=cos(3\pi)sin(sin(3\pi))=0$

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Calculus 1 : Differential Functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #832 : Other Differential Functions

Example Question #832 : How To Find Differential Functions

Example Question #834 : Other Differential Functions

Example Question #1021 : Differential Functions

Example Question #832 : Other Differential Functions

Example Question #1021 : Differential Functions

Example Question #837 : Other Differential Functions

Example Question #841 : Other Differential Functions

Example Question #1021 : Differential Functions

Example Question #841 : How To Find Differential Functions

All Calculus 1 Resources

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