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Calculus 1 : Differential Equations

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Example Questions

Example Question #341 : Equations

What is the derivative of $\ln(\cos(5x))$ at $\frac{\pi}{6}$ ?

Possible Answers:

$-\frac{1}{\sqrt3}$

$\frac{1}{\sqrt3}$

$-\frac{5}{\sqrt3}$

$\frac{\sqrt3}{5}$

$\frac{5}{\sqrt3}$

Correct answer:

$\frac{5}{\sqrt3}$

Explanation:

Here you must use the chain rule,

$\frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)$

and the rule for natural log,

$\frac{d}{dx}ln(x)=\frac{1}{x}$ .

First, you get

$\frac{(\cos(5x))')}{\cos(5x)}$ ,

then using the chain on $\cos(5x)$ , you get

$\frac{-\sin(5x)(5x)'}{\cos(5x)}$ .

You must then remember the , which gives you

$\frac{-5\sin(5x)}{\cos(5x)}$ as the final derivative.

Plugging in $\frac{\pi}{6}$ gives $\frac{5}{\sqrt3}$ .

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Example Question #343 : Equations

Find $\frac{dy}{dx}$ for the equation:

x^2+y^2=2xy

Possible Answers:

$\frac{2x-2y}{y^2}$

$\frac{x^2}{2x-2y}$

$\frac{2x-2y}{x^2}$

$\frac{y^2}{2x-2y}$

Correct answer:

Explanation:

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

x^2+y^2=2xy

The derivative is then

2xdx+0dy+0dx+2ydy=2ydx+2xdy

2xdx+2ydy=2ydx+2xdy

Now bring and terms to opposite sides of the equation:

(2x-2y)dx=(2x-2y)dy

Now rearraging variables gives $\frac{dy}{dx}$ :

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

$\frac{dy}{dx}=1$

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Example Question #2421 : Calculus

Find $\frac{dy}{dx}$ for the following equation:

cos(x^2)=3xy^2

Possible Answers:

$\frac{cos(2x)+3y}{xy}$

$\frac{6xy}{2xcos(x^2)-3y^2}$

$\frac{-2xsin(x^2)-3y^2}{6xy}$

$\frac{2xcos(x^2)}{6xy}$

$\frac{2xcos(x^2)}{3y}$

Correct answer:

$\frac{-2xsin(x^2)-3y^2}{6xy}$

Explanation:

Given the function:

cos(x^2)=3xy^2

We'll be taking the derivative with respect to the variables and .

Beginning with the left side of the equation, only the variable appears and the derivative is:

-2xsin(x^2)dx

Note that $d[cos(u)]=-dusin(u);d[x^n]=nx^{n-1}\rightarrow d[cos(x^n)]=-nx^{n-1}cos(x^n)$

Now for the right side of the equation, both and appear, so we'll utilize the chain rule giving the derivative:

3y^2dx+6xydy

Combining these gives the derivative of the original equation:

-2xsin(x^2)dx=3y^2dx+6xydy

Since we're looking for $\frac{dy}{dx}$ , gather terms on one side of the equation and terms on the other side:

(-2xsin(x^2)-3y^2)dx=(6xy)dy

From there, separate terms once more to find $\frac{dy}{dx}$ :

$\frac{-2xsin(x^2)-3y^2}{6xy}=\frac{dy}{dx}$

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Example Question #345 : Equations

Find $\frac{dx}{dy}$ for the equation:

$e^{xy}=xy-y^2$

Possible Answers:

$\frac{y(e^{xy}-1)}{x-2y-xe^{xy}}$

$\frac{y-2x-ye^{xy}}{y(e^{xy}-1)}$

$\frac{x-2y-xe^{xy}}{y(e^{xy}-1)}$

$\frac{x(e^{xy}-1)}{y-2x-ye^{xy}}$

$\frac{y-2x-ye^{xy}}{x(e^{xy}-1)}$

Correct answer:

$\frac{x-2y-xe^{xy}}{y(e^{xy}-1)}$

Explanation:

For this problem, knowledge of the following derivatives is necessary:

d[e^u]=du(e^u)

$d[x^n]=nx^{n-1};d[y^n]=ny^{n-1}$

To take the derivative of the equation

$e^{xy}=xy-y^2$

Let's begin with the left side. For each term, we'll take the derivative with respect to both variables and , treating the other variable as just a constant when we do so. The derivative of the left side is thus

$ye^{xy}dx+xe^{xy}dy$

Now moving to the right side, the derivative is:

ydx+xdy-2ydy

Notice how since the y^2 term has no term, when we take the derivative with respect to we just get zero, since we're treating the y^2 as a constant.

Now we have the derived equation:

$ye^{xy}dx+xe^{xy}dy=ydx+xdy-2ydy$

Bring and terms to opposite sides of the equation:

$(ye^{xy}-y)dx=(x-2y-xe^{xy})dy$

Now we can once more rearrange variables to find $\frac{dx}{dy}$ :

$\frac{dx}{dy}=\frac{x-2y-xe^{xy}}{y(e^{xy}-1)}$

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Example Question #346 : Equations

Find $\frac{dy}{dx}$ for the equation:

$e^{x}+y=2xy$

Possible Answers:

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

$\frac{e^{x}-2y}{2x-1}$

$\frac{2x-1}{e^{x}-2y}$

$\frac{ye^{x}}{2x-y}$

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

Correct answer:

$\frac{e^{x}-2y}{2x-1}$

Explanation:

We'll be taking the derivative of each term in the equation

$e^{x}+y=2xy$

with respect to both and . When taking the derivative for one variable, treat the other variable as a constant:

Note that d[e^u]=du(e^u)

$e^{x}dx+dy=2ydx+2xdy$

Notice how e^x is treated as a constant when taking the derivative with respect to and so goes to zero. The same happens when taking the derivative of with respect to

Now bring and terms to opposite sides of the equation:

$(e^{x}-2y)dx=(2x-1)dy$

Rearrange terms once more to find $\frac{dy}{dx}$ :

$\frac{e^{x}-2y}{2x-1}=\frac{dy}{dx}$

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Example Question #347 : Equations

Find $\frac{dy}{dx}$ for the equation:

$sin(xy^2)=3e^{2xy}$

Possible Answers:

$\frac{6e^{2xy}-2cos(xy^2)}{cos(xy^2)-6e^{2xy}}$

$\frac{cos(xy^2)-6xe^{2xy}}{6ye^{2xy}-2ycos(xy^2)}$

$\frac{6ye^{2xy}-2ycos(xy^2)}{cos(xy^2)-6xe^{2xy}}$

$\frac{y^2cos(xy^2)-6ye^{2xy}}{6xe^{2xy}-2xycos(xy^2)}$

$\frac{6xe^{2xy}-2xycos(xy^2)}{y^2cos(xy^2)-6ye^{2xy}}$

Correct answer:

$\frac{y^2cos(xy^2)-6ye^{2xy}}{6xe^{2xy}-2xycos(xy^2)}$

Explanation:

For this problem, note that:

d[sin(u)]=du(cos(u))

d[e^u]=du(e^u)

To approach this problem, we'll take the derivative of each term twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the equation

$sin(xy^2)=3e^{2xy}$

The derivative is then

$y^2cos(xy^2)dx+2xycos(xy^2)dy=6ye^{2xy}dx+6xe^{2xy}dy$

Now, bring and terms to opposite sides of the equation:

$(y^2cos(xy^2)-6ye^{2xy})dx=(6xe^{2xy}-2xycos(xy^2))dy$

Finally, rearrange terms once more to get $\frac{dy}{dx}$ :

$\frac{y^2cos(xy^2)-6ye^{2xy}}{6xe^{2xy}-2xycos(xy^2)}=\frac{dy}{dx}$

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Example Question #91 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

ln(xy)=x^2-2y

Possible Answers:

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

$\frac{2+\frac{x}{y}}{2-\frac{y}{x}}$

$\frac{2x-\frac{y}{x}}{2+\frac{x}{y}}$

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

$\frac{2+\frac{x}{y}}{2x-\frac{y}{x}}$

Correct answer:

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

Explanation:

For this problem, note that:

$d[ln(u)]=\frac{du}{u}$

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

ln(xy)=x^2-2y

The derivative of each side is

$\frac{y}{xy}dx+\frac{x}{xy}dy=2xdx+0dy-0dx-2dy$

$\frac{1}{x}dx+\frac{1}{y}dy=2xdx-2dy$

Now move and terms to opposite sides of the equation:

$(\frac{1}{x}-2x)dx=-(2+\frac{1}{y})dy$

Finally rearrange variables to get $\frac{dy}{dx}$ :

$\frac{2x-\frac{1}{x}}{2+\frac{1}{y}}=\frac{dy}{dx}$

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Example Question #92 : How To Find Solutions To Differential Equations

Find $\frac{dx}{dy}$ for the equation:

$ln(x^2y)=e^{xy}$

Possible Answers:

$\frac{\frac{2}{x}-ye^{xy}}{xe^{xy}-\frac{1}{y}}$

$\frac{\frac{2y}{x}-ye^{xy}}{xe^{xy}-\frac{x}{y}}$

$\frac{\frac{2}{x}-e^{xy}}{e^{xy}-\frac{1}{y}}$

$\frac{xe^{xy}-\frac{x}{y}}{\frac{2y}{x}-ye^{xy}}$

$\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

Correct answer:

$\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

Explanation:

For this problem, note that:

$d[ln(u)]=\frac{du}{u}$

d[e^u]=du(e^u)

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

$ln(x^2y)=e^{xy}$

The derivative is then

$\frac{2xy}{x^2y}dx+\frac{x^2}{x^2y}dy=ye^{xy}dx+xe^{xy}dy$

$\frac{2}{x}dx+\frac{1}{y}dy=ye^{xy}dx+xe^{xy}dy$

Now bring and terms to opposite sides of the equation:

$(\frac{2}{x}-ye^{xy})dx=(xe^{xy}-\frac{1}{y})dy$

Now rearraging variables gives $\frac{dx}{dy}$ :

$\frac{dx}{dy}=\frac{xe^{xy}-\frac{1}{y}}{\frac{2}{x}-ye^{xy}}$

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Example Question #93 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

sin(x^2y)cos(3xy^3)=1

Possible Answers:

$\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

$\frac{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}$

$\frac{xycos(x^2y)cos(3xy^3)-y^3sin(x^2y)sin(3xy^3)}{xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

$\frac{9xy^2sin(x^2y)sin(3xy^3)+x^2cos(x^2y)cos(3xy^3)}{2xycos(x^2y)cos(3xy^3)+3y^3sin(x^2y)sin(3xy^3)}$

$\frac{2xycos(x^2y)cos(3xy^3)+3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)+x^2cos(x^2y)cos(3xy^3)}$

Correct answer:

$\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}$

Explanation:

For this problem, note that:

d[sin(u)]=du(cos(u))

d[cos(u)]=-du(sin(u))

Product rule: d[uv]=udv+vdu

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

sin(x^2y)cos(3xy^3)=1

The derivative is then

${2xycos(x^2y)cos(3xy^3)dx-3y^3sin(x^2y)sin(3xy^3)dx+x^2cos(x^2y)cos(3xy^3)dy-9xy^2sin(x^2y)sin(3xy^3)dy=0}$

Remember to utilize the chain rule!

Now bring and terms to opposite sides of the equation:

${(2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3))dx=(9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3))dy}$

Now rearraging variables gives $\frac{dy}{dx}$ :

${\frac{2xycos(x^2y)cos(3xy^3)-3y^3sin(x^2y)sin(3xy^3)}{9xy^2sin(x^2y)sin(3xy^3)-x^2cos(x^2y)cos(3xy^3)}=\frac{dy}{dx}}$

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Example Question #94 : How To Find Solutions To Differential Equations

Find $\frac{dy}{dx}$ for the equation:

x^2sin(xy^2)=y

Possible Answers:

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1+2x^3ycos(xy^2)}$

$\frac{1+2x^3ycos(xy^2)}{x(2sin(xy^2)+xy^2cos(xy^2))}$

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}$

$\frac{x(sin(xy^2)+xy^2cos(xy^2))}{1+x^3ycos(xy^2)}$

$\frac{1-2x^3ycos(xy^2)}{x(2sin(xy^2)+xy^2cos(xy^2))}$

Correct answer:

$\frac{x(2sin(xy^2)+xy^2cos(xy^2))}{1-2x^3ycos(xy^2)}$

Explanation:

Note that:

d[sin(u)]=du(cos(u))

Product Rule: d[uv]=udv+vdu

Take the derivative of each term in the equation twice: with respect to and then with respect to . When taking the derivative with respect to one variable, treat the other variable as a constant.

For the function

x^2sin(xy^2)=y

The derivative is then found using the product rule to be:

2xsin(xy^2)dx+x^2y^2cos(xy^2)dx+2x^3ycos(xy^2)dy=dy

Notice how the chain rule needs to be utilized an additional time when taking the derivative of the x^2sin(xy^2) term with respect to .

Now bring and terms to opposite sides of the equation:

(2xsin(xy^2)+x^2y^2cos(xy^2))dx=(1-2x^3ycos(xy^2))dy

Now rearraging variables gives $\frac{dy}{dx}$ :

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

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Calculus 1 : Differential Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #341 : Equations

Example Question #343 : Equations

Example Question #2421 : Calculus

Example Question #345 : Equations

Example Question #346 : Equations

Example Question #347 : Equations

Example Question #91 : How To Find Solutions To Differential Equations

Example Question #92 : How To Find Solutions To Differential Equations

Example Question #93 : How To Find Solutions To Differential Equations

Example Question #94 : How To Find Solutions To Differential Equations

All Calculus 1 Resources

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