Equations - AP Calculus AB

Card 0 of 5558

Question

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

Answer

Note that:

Product Rule:

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

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

Notice how the chain rule needs to be utilized an additional time when taking the derivative of the term with respect to .

Now bring and terms to opposite sides of the equation:

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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Question

Find $f_{xyz}$ if $f(x,y,z)=e^{yz}cos(x^2z)$

Answer

For this problem, note that:

Product Rule

To solve this problem, differentiate the expression one variable at a time, treating other variables as constants:

If we're looking for $f_{xyz}$ for the function $f(x,y,z)=e^{yz}cos(x^2z)$ then we'll begin by differentiating with respect to first:

$f_x=-2xze^{yz}sin(x^2z)$

Next, differentiate with respect to :

$f_{xy}=-2xz^2e^{yz}sin(x^2z)$

Now finally we'll differentiate with respect to ; remember to use the product rule:

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xz^2(ye^{yz}sin(x^2z)+x^2e^{yz}cos(x^2z))$

$f_{xyz}=-4xze^{yz}sin(x^2z)-2xyz^2e^{yz}sin(x^2z)-2x^3z^2e^{yz}cos(x^2z))$

$f_{xyz}=-2xz(2e^{yz}sin(x^2z)+yze^{yz}sin(x^2z)+x^2ze^{yz}cos(x^2z)))$

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Question

Find given:

Answer

To solve, simply find the first derivative and let . Thus,

$f'(x)=2x-16\Rightarrow f'(-1)=2(-1)-16=-2-16=-18$

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Question

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

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

Answer

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

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

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:

Notice how since the term has no term, when we take the derivative with respect to we just get zero, since we're treating the 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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Question

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

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

Answer

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

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

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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Question

Find Dx\[sin(7x)\].

Answer

First, remember that Dx\[sin(x)\]=cos(x). Now we can solve the problem using the Chain Rule.

Dx\[sin(7x)\]

=cos(7x)Dx\[7x\]

=cos(7x)(7)

=7cos(7x)

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Question

Calculate fxxyz if f(x,y,z)=sin(4x+yz).

Answer

We can calculate this answer in steps. We start with differentiating in terms of the left most variable in "xxyz". So here we start by taking the derivative with respect to x.

First, fx= 4cos(4x+yz)

Then, fxx= -16sin(4x+yz)

fxxy= -16zcos(4x+yz)

Finally, fxxyz= -16cos(4x+yz) + 16yzsin(4x+yz)

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Question

Integrate $\small \int_{0}^{\pi } \cos x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sin x=cosx$

thus:

$\small \small \int_{0}^{\pi } \cos x \hspace 2 dx = \sin x \left]^\pi_0$
$\small \sin \pi - \sin 0 = 0 - 0 = 0$

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Question

Find the derivative of (5+3x)5.

Answer

We'll solve this using the chain rule.

Dx\[(5+3x)5\]

=5(5+3x)4 * Dx\[5+3x\]

=5(5+3x)4(3)

=15(5+3x)4

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Question

Integrate :

$\small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx$

Answer

$\frac{\mathrm{d} }{\mathrm{d} x}\sec x=\sec x\tan x$

thus:
$\small \small \int_{-\frac{\pi}{4}}^{0} \sec x \tan x \hspace 2 dx = \sec x \left]^0_{-\frac{\pi}{4} }$
$\small = \sec 0 - \sec(-\frac{\pi}{4}) = 1 - \sqrt{2}$

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Question

Find the general solution, , to the differential equation

$\frac{dy}{dx}=30-2x$ .

Answer

We can use separation of variables to solve this problem since all of the "y-terms" are on one side and all of the "x-terms" are on the other side. The equation can be written as $\frac{dy}{dx}=30-2x\Rightarrow dy=(30-2x)dx$ .

Integrating both sides gives us $\int dy = \int (30-2x) dx\Rightarrow y(x) = 30x-x^2+C$ .

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Question

Consider ; by multiplying by both the left and the right hand sides can be swiftly integrated as

$\frac{1}{2}(y')^2 = F(y) + C$

where . So, for example, can be rewritten as:

$\frac{1}{2} [y'(x)]^2 = \frac{1}{3} [y(x)]^3 + C$ . We will use this trick on another simple case with an exact integral.

Use the technique above to find such that with and .

Hint: Once you use the above to simplify the expression to the form , you can solve it by moving into the denominator:

$\int dy/g(y) = \int dx$

Answer

As described in the problem, we are given

.

We can multiply both sides by :

Recognize the pattern of the chain rule in two different ways:

$\frac{1}{2}[(y')^2]' = [y^2]' + \frac{1}{2} [y^4]'$

This yields:

We use the initial conditions to solve for C, noticing that at and This means that C must be 1 above, which makes the right hand side a perfect square:

$\frac{dy}{dx} = \pm (1 + y^2)$

To see whether the + or - symbol is to be used, we see that the derivative starts out positive, so the positive square root is to be used. Then following the hint we can rewrite it as:

$\int \frac{dy}{1 + y^2} = \int dx$ ,

which we learned to solve by the trigonometric substitution, yielding:

$\tan^{-1} y = x + C$

Clearly and the fact that again gives us so

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Question

What are all the functions such that

?

Answer

Integrating once, we get:

$y'(x) = \int \frac{dx}{x} + C_1 = \ln x + C_1$

Integrating a second time gives:

$y(x) = \int \ln x ~ dx + C_1 ~ x + C$

We integrate the first term by parts using to get:

$y(x) = x ~ \ln x ~-~ \int x ~ \frac{dx}{x} ~+~ C_1 ~ x ~+~ C$

Canceling the x's we get:

$y(x) = x ~ \ln x ~+~ \left( C_1 - 1 \right ) x ~+~ C$

Defining gives the above form.

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Question

The Fibonacci numbers are defined as

$F_0 = 0, F_1 = 1, F_n = F_{n-1} + F_{n-2}$

and are intimately tied to the golden ratios $\phi_\pm = (1 \pm \sqrt{5})/2$ , which solve the very similar equation

$\phi^n_\pm = \phi^{n - 1}_\pm + \phi^{n-2}_\pm$ .

The n'th derivatives of a function are defined as:

$y^{(0)}(x) = y(x), y^{(n)}(x) = \frac{d}{dx} y^{(n - 1)}(x)$

Find the Fibonacci function defined by:

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

$f^{(0)}(0) = 0, f^{(1)}(0) = 1$

whose derivatives at 0 are therefore the Fibonacci numbers.

Answer

To solve $f^{(n)} = f^{(n-1)} + f^{(n-2)}$ , we ignore of the derivatives to get simply:

This can be solved by assuming an exponential function $y = C e^{k x}$ , which turns this expression into

,

which is solved by $k = \phi_\pm$ . Our general solution must take the form:

$f(x) = C_+ ~ e^{x \phi_+} + C_- ~ e^{x \phi_-}$

Plugging in our initial conditions and , we get:

$C_+ \ \phi_+ ~+~ C_- \ \phi_- = 1$

$C_+ = 1/(\phi_+ - \phi_-) = 1/(\frac{1+\sqrt 5}{2} - \frac{1-\sqrt 5}{2}) = 1/\sqrt{5}$

Hence the answer is:

$f(x) = (e^{x \phi_+} - e^{x \phi_-}) / \sqrt{5}$

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Question

Find of the following equation:

Answer

First take the derivative and then solve when x=2.

To find the derivative use the power rule which states when,

the derivative is $f'(x)=nx^{n-1}$ .

Therefore the derivative of our function is:

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Question

Find for the following equation:

Answer

To find the derivative of this function we will need to use the product rule which states to multiply the first function by the derivative of the second function and add that to the product of the second function and the derivative of the first function. In other words,

$f'(x)=h(x)\cdot g'(x)+g(x)\cdot h'(x)$

To do this we will let,

and

and

Now we can find the derivative by plugging in these equations as follows.

Now plug in x=1 and solve.

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Question

Find the solution to the following equation at

Answer

To solve, we must first find the derivative and then solve when x=-2.

To find the derivative of the function we will use the Power Rule:

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Therefore,

Now to solve for -2 we plug it into our x value.

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Question

For:

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

Answer

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

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

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:

Where f(x) and g(x) are differentiable functions

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

In this case:

,

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 and , simply apply the chain rule:

For:

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

For:

$\frac{dy}{dx}=-sin(u)\frac{du}{dx}$ , 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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Question

Find for the following equation:

$f(x)=\ln(x)+7x^2-19x$

Answer

First, find the derivative. Then, evaluate at x=3.

For this function we will use the Power Rule to find the derivative.

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$

Also remember that the derivative of $\ln(x)$ is $\frac{1}{x}$ .

Therefore we get,

$f(x)=\ln(x)+7x^2-19x$

$f'(x)=\frac{1}{x}+14x-19$

$f'(3)=\frac{1}{3}+14(3)-19=23\frac{1}{3}$

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Question

Find the particular solution given .

$\frac{\mathrm{d} y}{\mathrm{d} x}=\frac{9x^2}{y^2-4y}$

Answer

The first thing we must do is rewrite the equation:

We can then find the integrals:

$\int (y^2-4y)dy= \int (9x^2)dx$

The integrals as as follows:

$\int (y^2-4y)dy= \frac{1}{3}y^3-2y^2$

$\int (9x^2)dx=3x^3$

we're left with

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

We then plug in the initial condition and solve for

$\frac{1}{3}(3)^3-2(3)^2=3(1)^3+c$

The particular solution is then:

$\frac{1}{3}y^3-2y^2=3x^3-12$

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