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Calculus 2 : First and Second Derivatives of Functions

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Example Question #51 : First And Second Derivatives Of Functions

Find the derivative of the function $y= \ln(\pi e^{15x})$

Possible Answers:

$15\pi\ln(e^{15x})$

None of the other answers

$15+\pi$

$\pi x \ln(e^{15x})$

Correct answer:

Explanation:

We could use the chain rule for this function, but we can save some time and work if we recognize that the function can be simplified a bit.

$y = \ln(\pi e^{15x})$

$= \ln(\pi) + \ln(e^{15x}})$

$= \ln(\pi) + 15x$

Hence

y' = 0 + 15 = 15

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Example Question #52 : First And Second Derivatives Of Functions

Evaluate $\frac{d}{dx}(xe^{\sin x})$

Possible Answers:

None of the other answers

$(x\cos x+1)e^{\sin x}$

$xe^{\sin x}$

$(xe^{\cos{x}}+1)x$

$(x\sin x +1)e^{\sin x}$

Correct answer:

$(x\cos x+1)e^{\sin x}$

Explanation:

To evaluate this derivative, we use the Product Rule.

$\frac{d}{dx}(xe^{\sin x})$

$= (x)(\cos xe^{\sin x})+(1)(e^{\sin x})$ . Use the Product Rule. Keep in mind that the derivative of $e^{\sin x}$ involves the Chain Rule.

$=(x\cos x +1)e^{\sin x}$ . Factor out an $e^{\sin x}$ .

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Example Question #53 : First And Second Derivatives Of Functions

Find the derivative of the following function:

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

Possible Answers:

$\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

$-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

$-\sin(3xe^{3x})(e^{3x}+9xe^{3x})$

$-\sin(3xe^{3x})(12e^{3x})$

Correct answer:

$-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

Explanation:

The derivative of the function is

$f'(x)=-\sin(3xe^{3x})(3e^{3x}+9xe^{3x})$

and was found using the following rules:

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

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Example Question #54 : First And Second Derivatives Of Functions

What is the second derivative of the following function:

$f(x)=ln(lnx)+e^{sinx}$

Possible Answers:

$f''(x)=ln(\frac{1}{x})+e^{sinx}$

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+e^{sinx}cos^{2}x-sinxe^{sinx}$

$f''(x)=\frac{1}{x^{2}ln^{2}x}+e^{sinx}$

$f''(x)=ln\frac{1}{x}+cosxe^{sinx}$

$f''(x)=\frac{1}{xlnx}+cosxe^{sinx}$

Correct answer:

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+e^{sinx}cos^{2}x-sinxe^{sinx}$

Explanation:

To solce this problem we use the chain rule.

Taking the first derviative we get:

$f'(x)=\frac{1}{lnx}(\frac{1}{x})+e^{sinx}cosx$ , which simplifies to $f'(x)=\frac{1}{xlnx}+cosxe^{sinx}$

To take the second derivative, we use a combination of the chain and product rules. To use the chain rule on the first term of the equation, we can re-write $\frac{1}{xlnx}$ as $(xlnx)^{-1}$ . Taking the second derivative, we get the following:

$f''(x)= -(xlnx)^{-2}[(x)(\frac{1}{x})+(1)lnx)]+[(cosx)e^{sinx}(cosx)-sinxe^{sinx}]$

Which simplifies to

$f''(x)=-\frac{1}{x^{2}ln^{2}x}(lnx+1)+cos^{2}xe^{sinx}-sinxe^{sinx}$

Which simplifies further to:

$f''(x)=\frac{-1}{x^{2}lnx}-\frac{1}{x^{2}ln^{2}x}+cos^{2}xe^{sinx}-sinxe^{sinx}$

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Example Question #55 : First And Second Derivatives Of Functions

Find the first derivative of the function:

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

Possible Answers:

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

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

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

$6xe^{3x^2}\sin(e^{3x^2})$

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

Correct answer:

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

Explanation:

The derivative of the function is equal to

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

and was found using the following rules:

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

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Example Question #56 : First And Second Derivatives Of Functions

If y=(3cos(x)sin(x))^2 , find $\frac{dy}{dx}$

Possible Answers:

$\frac{dy}{dx}=18$

$\frac{dy}{dx}=18sin(x)cos(2x)$

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

$\frac{dy}{dx}=18sin(2x)cos(2x)$

Correct answer:

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

Explanation:

By the chain rule:

$\frac{dy}{dx}=2(3cos(x)sin(x))*\frac{d}{dx}(3cos(x)sin(x))$

By the product rule:

$\frac{d}{dx}(3cos(x)sin(x))=3(cos(x)*cos(x)+sin(x)*-sin(x))$

=3cos^2(x)-3sin^2(x)=3cos(2x)

Therefore:

$\frac{dy}{dx}=18cos(x)sin(x)cos(2x)$

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Example Question #57 : First And Second Derivatives Of Functions

Find the derivative of f(x)=9x^2(4-3x^3) .

Possible Answers:

81x^4(4-3x^2)

18x^2-27x^4

-162x^3(4-3x^2)

36x^2-27x^5

72x-135x^4

Correct answer:

72x-135x^4

Explanation:

First, we should simplify the problem by distributing through the parenthesis.

9x^2(4-3x^3)=36x^2-27x^5 .

Now, since we have a polynomial, we use the power rule to take the derivative. Multiply the coefficient by the exponent, and reduce the power by 1.

f'(x)=36*2x^1-27*5x^4=72x-135x^4 .

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Example Question #58 : First And Second Derivatives Of Functions

Find $\frac{dy}{dx}$ using implicit differentiation of x^2 +y^2=25 .

Possible Answers:

$\frac{dy}{dx}=\frac{x}{2y}$

$\frac{dy}{dx}=\frac{x}{y}$

$\frac{dy}{dx}=\frac{-2x}{y}$

$\frac{dy}{dx}=\frac{-x}{y}$

$\frac{dy}{dx}=\frac{2x}{y}$

Correct answer:

$\frac{dy}{dx}=\frac{-x}{y}$

Explanation:

For implicit differentiation, you take a derivative of both the and components, and add a $\frac{dy}{dx}$ after every derivative. Then, using algebra, solve for the $\frac{dy}{dx}$ in the equation.

The derivative is: $2x+2y\frac{dy}{dx}=0$ , noting that the derivative of a constant equals zero. Now, we simply rearrange the equation.

$\frac{dy}{dx}=\frac{-2x}{2y}=\frac{-x}{y}.$

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Example Question #59 : First And Second Derivatives Of Functions

Find the derivative of ln(cos(3x^2+2)) .

Possible Answers:

-6xcot(3x^2+2)

6xtan(3x^2+2)

6xcot(3x^2+2)

-6xtan(3x^2+2)

Correct answer:

-6xtan(3x^2+2)

Explanation:

This derivative has multiple layers of the chain rule. Whenever woking with a chain rule derivative, always take the derivative of the outside function, leaving the inside function alone. Then, multiply that by the inside derivative. Here, our first outside function is ln(x) . The derivative of that function is $\frac{1}{x}$ . Then, we multiply that by the derivatives of the inside (which is another chain rule. The whole chain looks like this:

$ln(cos(3x^2+2))'=\frac{1}{cos(3x^2+2)}*-sin(3x^2+2)*6x$

$=\frac{-6xsin(3x^2+2)}{cos(3x^2+2)}=-6xtan(3x^2+2)$ .

In the last step, we used the definition $tanx=\frac{sinx}{cosx}$ to simplify the answer.

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Example Question #51 : First And Second Derivatives Of Functions

Find f'(x) .

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

Possible Answers:

$\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

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

$\frac{-10x+9x^2}{(3x^3-5x^2)^{2}}$

Correct answer:

$\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

Explanation:

To simplify the problem, it is easiest if we transform the function from in the denominator into the numinator.

$1/{(3x^3-5x^2)}=(3x^3-5x^2)^{-1}$ .

Now, we just take the derivative using the chain rule:

$(3x^3-5x^2)^{-1}'=-1(3x^3-5x^2)^{-2}*(9x^2-10x)$

$=\frac{10x-9x^2}{(3x^3-5x^2)^{2}}$

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Calculus 2 : First and Second Derivatives of Functions

Study concepts, example questions & explanations for Calculus 2

All Calculus 2 Resources

Example Questions

Example Question #51 : First And Second Derivatives Of Functions

Example Question #52 : First And Second Derivatives Of Functions

Example Question #53 : First And Second Derivatives Of Functions

Example Question #54 : First And Second Derivatives Of Functions

Example Question #55 : First And Second Derivatives Of Functions

Example Question #56 : First And Second Derivatives Of Functions

Example Question #57 : First And Second Derivatives Of Functions

Example Question #58 : First And Second Derivatives Of Functions

Example Question #59 : First And Second Derivatives Of Functions

Example Question #51 : First And Second Derivatives Of Functions

All Calculus 2 Resources

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