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Calculus 1 : How to find differential functions

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Example Questions

Example Question #358 : Differential Functions

Find the derivative of the function

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

Possible Answers:

$\frac{x^2e^{x^2}-3e^{x^2}}{x^4}$

$\frac{xe^{x^2}-3e^{x^2}}{x^4}$

$\frac{xe^{x^2}3e^{x^2}}{x^}$

$\frac{2x^2e^{x^2}-3e^{x^2}}{x^}$

$\frac{2x^2e^{x^2}-3e^{x^2}}{x^4}$

Correct answer:

$\frac{2x^2e^{x^2}-3e^{x^2}}{x^4}$

Explanation:

Since there is a division in the function:

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

It calls for the use of the quotient rule of derivatives:

$\frac{d}{dx}\frac{u}{v}=\frac{vdu-udv}{v^2}$

The derivative of $e^{x^2}$ requires use of the chain rule. This follows the form of:

$\frac{d}{dx}e^{g(x)}=\left(\frac{d}{dx}g(x)\right)\cdot e^{g(x)}$

$\frac{d}{dx}e^{x^2}=2xe^{x^2}$

The function in the quotient, x^3 , is of the form x^n whose derivative is $nx^{n-1}$ .

Therefore:

$\frac{d}{dx}x^3=3x^2$

Putting all of these things together, we can find the requested derivative:

$\frac{d}{dx}f(x)=\frac{x^3(2xe^{x^2})-e^{x^2}(3x^2)}{(x^3)^2}$

$\frac{d}{dx}f(x)=\frac{2x^4e^{x^2}-3x^2e^{x^2}}{x^6}$

$\frac{d}{dx}f(x)=\frac{2x^2e^{x^2}-3e^{x^2}}{x^4}$

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Example Question #171 : How To Find Differential Functions

Find the derivative of the function

$f(x)=sin(e^{cos(x^2)})$ .

Possible Answers:

$2xsin(x^2)e^{cos(x^2)}cos(e^{cos(x^2)})$

$-2xsin(x^2)e^{cos(x^2)}cos(e^{cos(x^2)})$

$-2xsin(x^2)e^{cos(x^2)}sin(e^{cos(x^2)})$

$2xsin(x^2)e^{cos(x^2)}sin(e^{cos(x^2)})$

$2xsin(x^2)e^{cos(x^2)}$

Correct answer:

$-2xsin(x^2)e^{cos(x^2)}cos(e^{cos(x^2)})$

Explanation:

This problem requires use of the chain rule.

$f(x)=sin(e^{cos(x^2)})$

Begin with the outside sin function; the derivative of a sin function follows the form:

$\frac{d}{dx}sin(g(x))=g'(x)cos(g(x))$ , where the ['] is used to desiginate a derivative.

In this case, the $g(x)=e^{cos(x^2)}$ .

The derivative of a natural log function follows the form:

$\frac{d}{dx}e^{h(x)}=h'(x)e^{h(x)}$

Here h(x)=cos(x^2)

And finally for a cosine function:

$\frac{d}{dx}cos(j(x))=-j'(x)sin(j(x))$

Where j(x)=x^2

So putting these all together:

j'(x)=2x

h'(x)=-2xsin(x^2)

$g'(x)=-2xsin(x^2)e^{cos(x^2)}$

$f'(x)=-2xsin(x^2)e^{cos(x^2)}cos(e^{cos(x^2)})$

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Example Question #171 : How To Find Differential Functions

Differentiate the function:

$f(t)=t^3\cos (t)$

Possible Answers:

$t^3+\cos (t)$

$3t^2-\sin (t)$

$3t^2\cos (t)-t^3\sin (t)$

$-3t^2\sin (t)$

Correct answer:

$3t^2\cos (t)-t^3\sin (t)$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dt}(t^3cos(t))\\ \\=t^3\frac{d}{dt}(cos(t))+(cos(t))\frac{d}{dt}(t^3)\\ \\=t^3(-sin(t))+cos(t)*3t^{3-1}\\ \\=-t^3sin(t)+3t^2cos(t)\\ \\=3t^2cos(t)-t^3sin(t)$

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Example Question #172 : How To Find Differential Functions

Differentiate the function:

$f(x)=\sqrt{x}sin(x)$

Possible Answers:

$\sqrt{x}+sin(x)$

$x^{\frac{1}{2}}cox(x)+\frac{1}{2}sin(x)x^{-\frac{1}{2}}$

$\frac{1}{2}x^{-\frac{1}{2}}cos(x)$

$\frac{1}{2}x^{-\frac{1}{2}}+cos(x)$

Correct answer:

$x^{\frac{1}{2}}cox(x)+\frac{1}{2}sin(x)x^{-\frac{1}{2}}$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dx}(\sqrt{x}sin(x))\\ \\=\frac{d}{dx}(x^{\frac{1}{2}}sin(x))\\ \\=x^{\frac{1}{2}}\frac{d}{dx}(sin(x))+sin(x)\frac{d}{dx}(x^{\frac{1}{2}})\\ \\=x^{\frac{1}{2}}cos(x)+sin(x)\frac{1}{2}x^{\frac{1}{2}-1}\\ \\=x^{\frac{1}{2}}cos(x)+\frac{1}{2}sin(x)x^{-\frac{1}{2}}$

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Example Question #172 : How To Find Differential Functions

Differentiate the function:

$f(x)=\left(\frac{1}{x^2}-\frac{3}{x^4}\right)(x+5x^3)$

Possible Answers:

$(-2x^{-3}+12x^{-5})+(1+15x^2)$

$(x^{-2}-3x^{-4})(1+15x^2)+(x+5x^3)(-2x^{-3}+12x^{-5})$

$\left(\frac{1}{x^2}-\frac{3}{x^4}\right)+(x+5x^2)$

$(-2x^{-3}+12x^{-5})(1+15x^2)$

Correct answer:

$(x^{-2}-3x^{-4})(1+15x^2)+(x+5x^3)(-2x^{-3}+12x^{-5})$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dx}\left(\left(\frac{1}{x^2}-\frac{3}{x^4}\right)(x+5x^3)\right)\\ \\=(x^{-2}-3x^{-4})\frac{d}{dx}(x+5x^3)+(x+5x^3)\frac{d}{dx}(x^{-2}-3x^{-4})\\ \\=(x^{-2}-3x^{-4})(1*x^{1-1}+3*5x^{3-1})+(x+5x^3)(-2x^{-2-1}-(-4)*3x^{-4-1})\\ \\=(x^{-2}-3x^{-4})(1+15x^2)+(x+5x^3)(-2x^{-3}+12x^{-5})$

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Example Question #173 : How To Find Differential Functions

Differentiate the following function:

$f(x)=(x^3-2x)(x^{-4}+x^{-2})$

Possible Answers:

$(x^3-2x)(-4x^{-5}-2x^{-3})+(x^{-4}+x^{-2})(3x^2-2)$

$(-4x^{-5}-2x^{-3})+(3x^2-2)$

$(x^3-2x)+(x^{-4}+x^{-2})$

$(-4x^{-5}-2x^{-3})(3x^2-2)$

Correct answer:

$(x^3-2x)(-4x^{-5}-2x^{-3})+(x^{-4}+x^{-2})(3x^2-2)$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dx}[(x^3-2x)(x^{-4}+x^{-2})]\\ \\=(x^3-2x)\frac{d}{dx}(x^{-4}+x^{-2})+(x^{-4}+x^{-2})\frac{d}{dx}(x^3-2x)\\ \\=(x^3-2x)((-4)x^{-4-1}+(-2)x^{-2-1})+(x^{-4}+x^{-2})(3x^{3-1}-2x^{1-1})\\ \\=(x^3-2x)(-4x^{-5}-2x^{-3})+(x^{-4}+x^{-2})(3x^2-2)$

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Example Question #173 : How To Find Differential Functions

Differentiate the function:

f(x)=sin(x)+cot(x)

Possible Answers:

cos(x)-csc(x)

cos(x)+csc^2(x)

-cos(x)-csc^2(x)

cos(x)-csc^2(x)

Correct answer:

cos(x)-csc^2(x)

Explanation:

Because this is a sum of functions, take the derivate of each function with respect to "x" and add them together:

$\\ \frac{d}{dx}[sin(x)+cot(x)]\\ \\=\frac{d}{dx}(sin(x))+\frac{d}{dx}(cot(x))\\ \\=cos(x)+(-csc^(x))\\ \\=cos(x)-csc^2(x)$

The derivative of sin(x) is cos(x) and the derivate of cot(x) is -csc^2(x) .

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Example Question #174 : How To Find Differential Functions

Differentiate the function:

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

Possible Answers:

$2x\cos (x)$

$x^2\cos (x)+2x\sin(x)$

$2x+\cos (x)$

Correct answer:

$x^2\cos (x)+2x\sin(x)$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dx}[x^2sin(x)]\\ \\=x^2\frac{d}{dx}(sin(x))+sin(x)\frac{d}{dx}(x^2)\\ \\=x^2cos(x)+sin(x)2x^{2-1}\\ \\=x^2cos(x)+2xsin(x)$

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Example Question #175 : How To Find Differential Functions

Differentiate the function:

$f(x)=(x^2+2)(x^{\frac{1}{2}}-1)$

Possible Answers:

$(x^2+2)+(x^{\frac{1}{2}}-1)$

$2x+x^{-\frac{1}{2}}$

$\frac{5}{2}x^{\frac{3}{2}}-2x+x^{-\frac{1}{2}}$

$2x^{\frac{1}{2}}$

Correct answer:

$\frac{5}{2}x^{\frac{3}{2}}-2x+x^{-\frac{1}{2}}$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Then apply the product rule $nx^{n-1}$ (where "n" is the exponent) where needed.

$\\ \frac{d}{dx}[(x^2+2)(x^{\frac{1}{2}}-1)]\\ \\=(x^2+2)\frac{d}{dx}(x^{\frac{1}{2}}-1)+(x^{\frac{1}{2}}-1)\frac{d}{dx}(x^2+2)\\ \\=(x^2+2)(\frac{1}{2}x^{\frac{1}{2}-1}-0)+(x^{\frac{1}{2}}-1)(2x^{2-1}+0)\\ \\=\frac{1}{2}x^{-\frac{1}{2}}(x^2+2)+2x(x^{\frac{1}{2}}-1)\\ \\=\frac{1}{2}x^{\frac{3}{2}}+x^{-\frac{1}{2}}+2x^{\frac{3}{2}}-2x\\ \\=\frac{5}{2}x^{\frac{3}{2}}-2x+x^{-\frac{1}{2}}$

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Example Question #362 : Differential Functions

Differentiate the function:

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

Possible Answers:

$\cos (x)+\sin (x)$

$\cos^2(x)-\sin^2(x)$

$\cos (x)-\sin (x)$

$-\sin (x)\cos (x)$

Correct answer:

$\cos^2(x)-\sin^2(x)$

Explanation:

Apply the product rule:

$\frac{d}{dx}[f(x)*g(x)]=f(x)\frac{d}{dx}[g(x)]+g(x)\frac{d}{dx}[f(x)]$

Let f(x)=sin(x) and g(x)=cos(x)

$\\ \frac{d}{dx}[\sin (x)\cos (x)]\\ \\=\sin (x)\frac{d}{dx}\cos (x)+\cos (x)\frac{d}{dx}\sin (x)\\ \\=-\sin (x)\sin (x)+\cos (x)\cos (x)\\ \\=\cos^2(x)-\sin^2(x)$

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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 #358 : Differential Functions

Example Question #171 : How To Find Differential Functions

Example Question #171 : How To Find Differential Functions

Example Question #172 : How To Find Differential Functions

Example Question #172 : How To Find Differential Functions

Example Question #173 : How To Find Differential Functions

Example Question #173 : How To Find Differential Functions

Example Question #174 : How To Find Differential Functions

Example Question #175 : How To Find Differential Functions

Example Question #362 : Differential Functions

All Calculus 1 Resources

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