Calculus 2 : Derivatives

Study concepts, example questions & explanations for Calculus 2

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Example Questions

Example Question #91 : First And Second Derivatives Of Functions

Find the derivative of the function:

\(\displaystyle f(x)=\frac{1}{x}+x^2+5x+e^{\tan(x)}\)

Possible Answers:

\(\displaystyle -\frac{1}{x^2}+2x+5+\sec^2(x)(e^{\tan(x)})\)

\(\displaystyle \frac{1}{x^2}+2x+5+\sec^2(x)(e^{\tan(x)})\)

\(\displaystyle -\frac{1}{x^2}+2x+5+\sec(x)(e^{\tan(x)})\)

\(\displaystyle -\frac{1}{x^2}+7x+\sec^2(x)(e^{\tan(x)})\)

\(\displaystyle -\frac{1}{x^2}+2x+5+e^{\tan(x)}\)

Correct answer:

\(\displaystyle -\frac{1}{x^2}+2x+5+\sec^2(x)(e^{\tan(x)})\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=-\frac{1}{x^2}+2x+5+\sec^2(x)(e^{\tan(x)})\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\)

Example Question #291 : Derivative Review

Find the derivative of the function:

\(\displaystyle f(x)=\tan(x^2)+x^3\)

Possible Answers:

\(\displaystyle \sec^2(x^2)\tan(x^2)+3x^2\)

\(\displaystyle \sec^2(x^2)+3x^2\)

\(\displaystyle 2x\sec^2(x^2)+3x^2\)

\(\displaystyle 2x\sec(x^2)+3x^2\)

Correct answer:

\(\displaystyle 2x\sec^2(x^2)+3x^2\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=2x\sec^2(x^2)+3x^2\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)

Example Question #92 : First And Second Derivatives Of Functions

Find the derivative of the function:

\(\displaystyle f(x)=x^2\sin(2x^4)\)

Possible Answers:

\(\displaystyle 2x\sin(2x^4)-8x^5\cos(2x^4)\)

\(\displaystyle 2x\sin(2x^4)+8x^6\cos(2x^4)\)

\(\displaystyle 2x\sin(2x^4)+8x^5\cos(2x^4)\)

\(\displaystyle x^2\sin(2x^4)+8x^5\cos(2x^4)\)

Correct answer:

\(\displaystyle 2x\sin(2x^4)+8x^5\cos(2x^4)\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=2x\sin(2x^4)+8x^5\cos(2x^4)\)

and was found using the following rules:

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

Example Question #93 : First And Second Derivatives Of Functions

Find the derivative of the function:

\(\displaystyle f(x)=e^{3x^2}\cos(3x)\)

Possible Answers:

\(\displaystyle 6xe^{3x^2}\cos(3x)+3e^{3x^2}\sin(3x)\)

\(\displaystyle 3x^2e^{3x^2}\cos(3x)-3e^{3x^2}\sin(3x)\)

\(\displaystyle 6xe^{3x^2}\cos(3x)-3e^{3x^2}\sin(3x)\)

\(\displaystyle 6xe^{3x^2}\cos(3x)-e^{3x^2}\sin(3x)\)

\(\displaystyle 6x\cos(3x)-3e^{3x^2}\sin(3x)\)

Correct answer:

\(\displaystyle 6xe^{3x^2}\cos(3x)-3e^{3x^2}\sin(3x)\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=6xe^{3x^2}\cos(3x)-3e^{3x^2}\sin(3x)\)

and was found using the following rules:

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

Example Question #94 : First And Second Derivatives Of Functions

Find the derivative of the function:

\(\displaystyle f(x)=x^2+xe^{2x}+10x\)

Possible Answers:

\(\displaystyle e^{2x}(x+1)+2x+10\)

\(\displaystyle e^{2x}(2x+1)+2x+10\)

\(\displaystyle e^{2x}(2x+1)+12x\)

\(\displaystyle 2xe^{2x}+2x+10\)

Correct answer:

\(\displaystyle e^{2x}(2x+1)+2x+10\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=e^{2x}(2x+1)+2x+10\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\)

Example Question #291 : Derivatives

Find the derivative of the function:

\(\displaystyle f(x)=\sqrt{\cos(x^2)}\)

Possible Answers:

\(\displaystyle \frac{{-x\sin(x^2)}{\sqrt{\cos(x^2)}}}{2}\)

\(\displaystyle \frac{x\sin(x^2)}{\sqrt{\cos(x^2)}}\)

\(\displaystyle \frac{-x\sin(x^2)}{2\sqrt{\cos(x^2)}}\)

\(\displaystyle \frac{-x\sin(x^2)}{\sqrt{\cos(x^2)}}\)

Correct answer:

\(\displaystyle \frac{-x\sin(x^2)}{\sqrt{\cos(x^2)}}\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=\frac{-x\sin(x^2)}{\sqrt{\cos(x^2)}}\)

and was found using the following rules:

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

Note that the square root itself is the "outer" function when using the first rule, the chain rule.

Example Question #292 : Derivatives

Find the derivative of the function:

\(\displaystyle f(x)=\frac{x}{x+1}+3x\tan(x)\)

Possible Answers:

\(\displaystyle \frac{1}{(x+1)^2}+3x(\tan(x)+\sec^2(x))\)

\(\displaystyle \frac{2x+1}{(x+1)^2}+3\tan(x)+3x\sec^2(x)\)

\(\displaystyle \frac{1}{(x+1)^2}+3\tan(x)+3x\sec(x)\)

\(\displaystyle \frac{1}{(x+1)^2}+3\tan(x)+3x\sec^2(x)\)

Correct answer:

\(\displaystyle \frac{1}{(x+1)^2}+3\tan(x)+3x\sec^2(x)\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=\frac{1}{(x+1)^2}+3\tan(x)+3x\sec^2(x)\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \frac{f(x)}{g(x)}=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(x)g(x)=f'(x)g(x)+f(x)g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \tan(x)=\sec^2(x)\)

Example Question #293 : Derivatives

Find the second derivative of the function:

\(\displaystyle f(x)=e^{x^2}+\ln(x^2)\)

Possible Answers:

\(\displaystyle e^{x^2}+\frac{2}{x}\)

\(\displaystyle 2xe^{x^2}+\frac{2}{x}\)

\(\displaystyle 2xe^{x^2}+\frac{1}{x}\)

\(\displaystyle 2xe^{x^2}+\left |\frac{2}{x} \right |\)

Correct answer:

\(\displaystyle 2xe^{x^2}+\frac{2}{x}\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=2xe^{x^2}+\frac{2}{x}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{d} u}{\mathrm{d} x}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \ln(u)=\frac{1}{u}\frac{\mathrm{d} u}{\mathrm{d} x}\)

Example Question #294 : Derivatives

Find the derivative of the function:

\(\displaystyle f(x)=\sqrt[3]{x}+\sqrt{e^{x^2}}\)

Possible Answers:

\(\displaystyle \frac{1}{3\sqrt[3]{x}}+x\sqrt{e^{x^{2}}}\)

\(\displaystyle \frac{1}{3\sqrt[3]{x^2}}+x\sqrt{e^{x^{2}}}\)

\(\displaystyle \frac{1}{3\sqrt[3]{x^2}}+xe^x\)

\(\displaystyle \frac{1}{3\sqrt[3]{x^2}}+\frac{x\sqrt{e^{x^{2}}}}{2}\)

Correct answer:

\(\displaystyle \frac{1}{3\sqrt[3]{x^2}}+x\sqrt{e^{x^{2}}}\)

Explanation:

The derivative of the function is equal to

\(\displaystyle f'(x)=\frac{1}{3\sqrt[3]{x^2}}+x\sqrt{e^{x^{2}}}\)

and was found using the following rules:

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} f(g(x))=f'(g(x))\cdot g'(x)\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} x^n=nx^{n-1}\)\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} e^u=e^u\frac{\mathrm{du} }{\mathrm{d} x}\)

Note that all of the radicals act as "outer" functions when using the first rule, the chain rule.

Example Question #295 : Derivatives

Find the derivative of the function:

\(\displaystyle f(x)=\cos(4x)\sec(2x)\)

Possible Answers:

\(\displaystyle -4\sin(4x)\sec(2x)+\cos(4x)\sec(2x)\tan(2x)\)

\(\displaystyle 8\)

\(\displaystyle 0\)

\(\displaystyle -2\sin(2x)\)

\(\displaystyle 2\sin(2x)\)

Correct answer:

\(\displaystyle -4\sin(4x)\sec(2x)+\cos(4x)\sec(2x)\tan(2x)\)

Explanation:

The derivative of the function is equal to 

\(\displaystyle f'(x)=-4\sin(4x)\sec(2x)+2\cos(4x)\sec(2x)\tan(2x)\)

and was found using the following rules:

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

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