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Calculus 1 : Equations

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10 Diagnostic Tests 438 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #73 : Differential Equations

Find the derivative of the following function.

$f(x)=(3x^3+5x^2-7)^{^{10}}$

Possible Answers:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(3x^3+5x^2-7)^{9}$

$(9x^2+10x)(300x^3+50x^2-70)^{9}$

$10(3x^3+5x^2-7)^{9}$

$(90x^2-100x)(3x^3+5x^2-7)^{9}$

Correct answer:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

Explanation:

$D_x[(3x^3+5x^2-7)^{10}]$ Finding the derivative requires use of the chain rule.

$D_x[(f(x))^a]=a*f(x)^{a-1}*f'(x)$ Where f(x) is a differentiable function and a is an integer.

From the problem statement:

f(x)=3x^3+5x^2-7

Computing the derivative:

f'(x)=9x^2+10x

Applying the Chain Rule:

$D_x[(3x^3+5x^2-7)^{10}]=10*(3x^3+5x^2-7)^{9}*(9x^2+10x)$

Simplify to match the answer choice:

$(90x^2+100x)(3x^3+5x^2-7)^{9}$

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Example Question #68 : How To Find Solutions To Differential Equations

For:

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

Find $\frac{dy}{dx}$ :

Possible Answers:

2xsin(x-3)cos(x^2+2)-cos(x-3)sin(x^2+2)

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

2xsin(x-3)sin(x^2+2)-cos(x-3)cos(x^2+2)

2xcos(x^2+2)-sin(x^2+2)

2xcos(x+3)cos(x^2+2)-sin(x+3)sin(x^2+2)

Correct answer:

2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)

Explanation:

Computation of the derivative requires use of the Product Rule and Chain Rule.

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

$\frac{dy}{dx}=\frac{d}{dx}[sin(x^2+2)cos(x-3)]$

A good way to remember the Product Rule is by memorizing this saying: "First Times the Derivative of the Second, Plus the Second Times the Derivative of the First." Or if that doesn't help then you can just write out the formula:

For:

y=f(x)*g(x) Where f(x) and g(x) are differentiable functions

As you can see, the "saying" from above matches the formula.

In this case:

f(x)=sin(x^2+2) , g(x)=cos(x-3)

Applying the Product Rule:

$\frac{dy}{dx}=sin(x^2+2)\frac{d}{dx}[cos(x-3)]+cos(x-3)\frac{d}{dx}[sin(x^2+2)]$

To compute the derivatives of cos(x-3) and sin(x^2+2) , simply apply the chain rule:

For:

y=sin(u)

$\frac{dy}{dx}=cos(u)\frac{du}{dx}$ , Where u is a differentiable function

For:

y=cos(u)

$\frac{dy}{dx}=-sin(u)\frac{du}{dx}$ , Where u is a differentiable function

Applying the Chain Rule:

$\frac{dy}{dx}=sin(x^2+2)(-sin(x-3)(1))+cos(x-3)(cos(x^2+2)(2x))$

Simplify the expression to match one of the answer choices:

$\frac{dy}{dx}=2xcos(x-3)cos(x^2+2)-sin(x-3)sin(x^2+2)$

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Example Question #69 : How To Find Solutions To Differential Equations

Find the derivative with respect to x for the following function:

$y=e^{sin(3x)}+(x^5-3x)^3$

Possible Answers:

$\frac{dy}{dx}=3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3cos(3x)e^{sin(3x)}+(12x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

$\frac{dy}{dx}=-3sin(3x)e^{sin(3x)-1}+(12x^4-9)(x^5-3x^2)^2$

$\frac{dy}{dx}=3cos(3x)e^{sin(x)}+3(x^5-3x)^2$

Correct answer:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

Explanation:

Computation of the derivative requires use of the Chain Rule.

$y=e^{sin(3x)}+(x^5-3x)^3$

$\frac{dy}{dx}=\frac{d}{dx}[e^{sin(3x)}+(x^5-3x)^3)]$

For:

$y=e^{f(u)}$

$\frac{dy}{dx}=e^{f(u)}*f'(u)$

Here f(u)=sin(3x) , therefore, f'(u)=3cos(3x)

For:

y=(f(x))^a

$\frac{dy}{dx}=a*(f(x))^{a-1}*f'(x)$

Applying these two rules results in:

$\frac{dy}{dx}=e^{sin(3x)}(3cos(3x))+3(x^5-3x)^2(5x^4-3)$

Simplifying this equation results in one of the answer choices:

$\frac{dy}{dx}=3cos(3x)e^{sin(3x)}+(15x^4-9)(x^5-3x)^2$

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Example Question #1371 : Functions

Solve the differential equation:

$\frac{dy}{dx}=\frac{2sinx+cosx}{y}$

Possible Answers:

$y=\sqrt{2sinx-4cosx+c}$

$y=\sqrt{4cosx-2sinx+c}$

$y=\sqrt{8cosx-4sinx+c}$

y=4cosx-2sinx+c

Correct answer:

$y=\sqrt{2sinx-4cosx+c}$

Explanation:

This is a separable equation, meaning to solve we want to separate, then integrate.

Step 1: Separate by putting the y components on one side of the equation, and x components on the other.

ydy=(2sinx+cosx)dx

Step 2: Integrate both sides of the equation.

$\int ydy= \int (2sinx+cosx)dx$

This gives us the following:

$\frac{1}{2}y^{2}=-2cosx+sinx+c$

Step 3: Now to simplify we can solve for y.

$y^{2}=2sinx-4cosx+c$

$y=\sqrt{2sinx-4cosx+c}$

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Example Question #71 : How To Find Solutions To Differential Equations

Find the general solution for the following differential equation:

$\frac{dy}{dx}= x^{2}-2x+6$

Possible Answers:

$y = \frac{1}{3}x^{3} -\frac{1}{2}x^{2}+6x+c$

$y = \frac{1}{3}x^{3} -x^{2}+6x+c$

$y = 3x^{3} -x^{2}+6+c$

$y = x^{3} -2x^{2}+6x+c$

Correct answer:

$y = \frac{1}{3}x^{3} -x^{2}+6x+c$

Explanation:

This is separable differential equation. The first step is to get all of the x components of the function on one side and the y components on the other. Once that is done, integrate each side as shown:

$dy= (x^{2}-2x+6)dx \Rightarrow \int dy=\int x^{2}-2x+6$

To evaluate the integral, use the inverse power rule

$\int x^{n}dx = \frac{x^{n+1}}{n+1} n\neq1$ .

Integrating the function should result in the answer below:

$y = \frac{1}{3}x^{3} -x^{2}+6x+c$

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Example Question #71 : How To Find Solutions To Differential Equations

Solve the following differential equation

$\frac{dy}{dt}+3y=0$

y(0)=1

Possible Answers:

$y=e^{3t}$

$y=e^{-3t}$

$y=3e^{t}$

y=0

Correct answer:

$y=e^{-3t}$

Explanation:

Given the linear, homogenous differential equation

$\frac{dy}{dt}+3y=0$

The general solution is given by

$y=Ce^{-3t}$

Since y(0)=1 , C=1

The specific solution is therefore

$y=e^{-3t}$

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Example Question #72 : How To Find Solutions To Differential Equations

Find the derivative of the function.

$f(x)=e^{5x}\sin (4x)$

Possible Answers:

$5e^{5x}\sin (4x)+4e^{5x}\cos (4x)$

None of these

$5e^{x}\sin (4x)+4e^{x}\cos (4x)$

$e^{5x}\sin (4x)+4e^{5x}\cos (4x)$

$5e^{5x}\sin (4x)+e^{5x}\cos (4x)$

Correct answer:

$5e^{5x}\sin (4x)+4e^{5x}\cos (4x)$

Explanation:

To find the derivative of this function we must use the product rule.

It states that the derivative of f(x)g(x) is f'(x)g(x)+f(x)g'(x) .

We must also recognize that the derivative of sine is cosine and the derivative of e^x is e^x .

The derivative of this function is then

$5e^{5x}\sin (4x)+4e^{5x}\cos (4x)$ .

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Example Question #74 : How To Find Solutions To Differential Equations

Find the derivative of the function.

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

Possible Answers:

$\frac{1-3ln(3x)}{x^6}$

$\frac{1-3ln(3x)}{x^4}$

$\frac{1-ln(3x)}{x^4}$

$\frac{3-3ln(3x)}{x^4}$

None of these

Correct answer:

$\frac{1-3ln(3x)}{x^4}$

Explanation:

To find the derivative of of this function we must use the quotient rule.

It states that the derivative of

$\frac{f(x)}{g(x)}$ is $\frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}$ .

We must also use the rules that the derivative of ln(x) is $\frac{1}{x}$ and that the derivative of x^n is $nx^{n-1}$ .

Thus the derivative of the function is

$\frac{\frac{3}{3x}x^3-3x^2ln(3x)}{x^6}=\frac{x^2-3x^2ln(3x)}{x^6}=\frac{1-3ln(3x)}{x^4}$ .

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Example Question #73 : How To Find Solutions To Differential Equations

Find the derivative of the function.

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

Possible Answers:

$\frac{2cos(ln(x^2))}{x}$

$\frac{2cos(ln(x))}{x}$

None of these

$\frac{2cos(ln(x^2))}{x^2}$

$\frac{cos(ln(x^2))}{x}$

Correct answer:

$\frac{2cos(ln(x^2))}{x}$

Explanation:

To calculate this derivative we must use the chain rule.

The chain rule states that the derivative of f(g(x)) is f'(g(x))g'(x) .

We must also understand that the derivative of sine is cosine, the derivative of ln(x) is $\frac{1}{x}$ , and the derivative of x^n is $nx^{n-1}$ .

f(x)=sin(ln(x^2)) so we must take the derivative of the inner function which is ln(x^2) .

We must also use the chain rule for this.

The derivative of ln(x^2) is $\frac{2x}{x^2}=\frac{2}{x}$ .

The derivative of the outer function is cos(ln(x^2)) .

Thus the final answer is

$\frac{2cos(ln(x^2))}{x}$ .

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Example Question #74 : How To Find Solutions To Differential Equations

Determine possible solution to the differential equation

$\frac{d^2 y}{d^2x}+y=0$ .

Possible Answers:

y=tan(x)

$y=e^{t}*sin(t)$

y=cos(x) and y=sin(x)

$y=-e^{-t}$ and $y=e^{-t}$

Correct answer:

y=cos(x) and y=sin(x)

Explanation:

To solve this equation, notice that the second derivative of a function is equal to the negative of the original function.

$\frac{\mathrm{d^2}y }{\mathrm{d} x^2}=-y$

The only elementary functions that satisfy this rule are the sine and cosine functions.

Therefore, possible solutions to this differential equation is y=sin(x) and y=cos(x) .

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #73 : Differential Equations

Example Question #68 : How To Find Solutions To Differential Equations

Example Question #69 : How To Find Solutions To Differential Equations

Example Question #1371 : Functions

Example Question #71 : How To Find Solutions To Differential Equations

Example Question #71 : How To Find Solutions To Differential Equations

Example Question #72 : How To Find Solutions To Differential Equations

Example Question #74 : How To Find Solutions To Differential Equations

Example Question #73 : How To Find Solutions To Differential Equations

Example Question #74 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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