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Calculus 1 : Equations

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Example Questions

Example Question #361 : Equations

Find the derivative of the function.

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

Possible Answers:

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

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

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

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

None of these

Correct answer:

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

Explanation:

To find the derivative of this function we need to use the chain rule and multiplication rule. The chain rule states that the derivative of f(g(x)) is f'(g(x))g'(x) . The multiplication rule states that the derivative of f(x)g(x) is f'(x)g(x)+f(x)g'(x) . The derivative of x^n is $nx^{n-1}$ . The derivative of sin is cos and the derivative of cos is -sin. So lets say

g(x)=x then g'(x)=1 and

$h(x)=\sin (\cos (x))$ then $h'(x)=-\cos(\cos (x))\cdot\sin (x)$

So the answer is

$f'(x)=-x\cos (\cos (x))\cdot\sin (x)+\sin (\cos (x))$

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Example Question #2441 : Calculus

What is the slope of the function f(x,y)=sin(sin(xy^2)) at the point $(\pi,0)$ ?

Possible Answers:

$(\pi,0)$

$(0,\pi)$

(0,0)

$(0,\pi^2)$

$(\pi^2,0)$

Correct answer:

$(\pi^2,0)$

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Trigonometric derivative:

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

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of f(x,y)=sin(sin(xy^2)) at the point $(\pi,0)$

$\frac{\delta f}{\delta x}=y^2cos(xy^2)cos(sin(xy^2));\pi^2cos((0)\pi^2)cos(sin((0)\pi^2))=\pi^2$

$\frac{\delta f}{\delta y}=2xycos(xy^2)cos(sin(xy^2));2(0)(\pi)cos((0)\pi^2)cos(sin((0)\pi^2))=0$

The slope is $(\pi^2,0)$

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Example Question #102 : How To Find Solutions To Differential Equations

Find the slope of the function $f(x,y)=x(\frac{1}{2}^{4y})$ at the point (4,1) .

Possible Answers:

(-0.0625,-0.6931)

(0.0625,-0.6931)

(0.0625,0.6931)

(0.6931,0.0625)

(0.6931,-0.0625)

Correct answer:

(0.0625,-0.6931)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential:

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of $f(x,y)=x(\frac{1}{2}^{4y})$ at the point (4,1)

$\frac{\delta f}{\delta x}=\frac{1}{2}^{4y};\frac{1}{2}^{4(1)}=0.0625$

$\frac{\delta f}{\delta y}=4x(\frac{1}{2}^{4y})ln(\frac{1}{2});4(4)(\frac{1}{2}^{4(1)})ln(\frac{1}{2})=-0.6931$

The slope is (0.0625,-0.6931)

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Example Question #362 : Equations

Find the slope of the function $f(x,y)=\frac{1}{4}^{\frac{1}{3}xy}$ at the point (2,4) .

Possible Answers:

(-0.046,0.023)

(-0.046,-0.023)

(-0.023,-0.046)

(0.023,0.046)

(0.046,0.023)

Correct answer:

(-0.046,-0.023)

Explanation:

To consider finding the slope, let's discuss the topic of the gradient.

For a function f(x_1,x_2,...,x_n) , the gradient is the sum of the derivatives with respect to each variable, multiplied by a directional vector:

$\frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}$

It is essentially the slope of a multi-dimensional function at any given point

Knowledge of the following derivative rule will be necessary:

Derivative of an exponential:

d[a^u]=a^uduln(a)

Note that u may represent large functions, and not just individual variables!

The approach to take with this problem is to simply take the derivatives one at a time. When deriving for one particular variable, treat the other variables as constant.

Take the partial derivatives of $f(x,y)=\frac{1}{4}^{\frac{1}{3}xy}$ at the point (2,4)

$\frac{\delta f}{\delta x}=\frac{y}{3}(\frac{1}{4}^{\frac{1}{3}xy})ln(\frac{1}{4});\frac{4}{3}(\frac{1}{4}^{\frac{1}{3}(2)(4)})ln(\frac{1}{4})=-0.046$

$\frac{\delta f}{\delta y}=\frac{x}{3}(\frac{1}{4}^{\frac{1}{3}xy})ln(\frac{1}{4});\frac{2}{3}(\frac{1}{4}^{\frac{1}{3}(2)(4)})ln(\frac{1}{4})=-0.023$

The slope is (-0.046,-0.023) .

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Example Question #2444 : Calculus

Find the explicit function of given

$\frac{dy}{dx}=xy$ .

Possible Answers:

$y=Ce^{\frac{x^2}{2}}$

y=e^2+c

y=Ce^x

$y=e^{x^2}$

Correct answer:

$y=Ce^{\frac{x^2}{2}}$

Explanation:

In order to determine the explicit function of y, we must separate the variables onto each side of the equation

becomes

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

Integrating both sides of the equation

and by applying the inverse power rule for the right-hand-side which says

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

yields

$ln|y|=\frac{x^2}{2}+c$

Exponentiating both sides of the equation, we obtain

$y=Ce^{\frac{x^2}{2}}$

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Example Question #112 : How To Find Solutions To Differential Equations

Find the derivative of $f(s) = 2se^{2s}$ .

Possible Answers:

$2e^{2s} + 2se^{2s}$

$2e^{2s}$

$2se^{2s}$

$2e^{2s} + 4se^{2s}$

$4e^{2s}$

Correct answer:

$2e^{2s} + 4se^{2s}$

Explanation:

This function is composed of two functions multiplied together; therefore you must use the product rule to find the derivative. The product rule is given by:

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

Note in this case the two functions:

f(x) = 2s

$g(x) = e^{2s}$

The derivatives are:

f'(x) = 2

$g'(x) = 2e^{2s}$

Using the derivatives of the two functions and applying the product rule, you will recieve the proper derivative:

$2e^{2s} + 4se^{2s}$

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Example Question #113 : How To Find Solutions To Differential Equations

Which differential equations does y=ln(x) solve? Assume x>0 .

Possible Answers:

y'=e^y

$y'=\frac{1}{y}$

$y'=\frac{1}{e^y}$

$y=\frac{1}{e^y'}$

Correct answer:

$y'=\frac{1}{e^y}$

Explanation:

You can solve this equation by plugging in y=ln(x) into each answer choice and seeing if the two sides are equal.

Another method is to arrange in terms of itself and its derivative.

y=ln(x) , x>0

$y'=\frac{1}{x}$ . This is an identity.

We want to create a differential equation that equates those two terms.

y=ln(x)

Raising both sides by

$e^y=e^{ln(x)}$ .

Recall that $e^{ln(x)}=x$

e^y=x .

Raising both sides to the negative first power,

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

$\frac{1}{e^y}=\frac{1}{x}$

Recall that $\frac{1}{x}=y'$

Therefore,

$y'=\frac{1}{e^y}$

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Example Question #364 : Equations

Find the general solution to the differential equation given by:

y''=3t

Assume is a function of . and given below are constants.

Possible Answers:

$y=\frac{1}{2}t^3+Ct$

$y=\frac{1}{2}t^3+C$

$y=\frac{1}{2}t^3+Bt+C$

y=0

Correct answer:

$y=\frac{1}{2}t^3+Bt+C$

Explanation:

To solve this, we only have to take the integral of both sides twice, and that will remove the y'' term.

Remember the power rule for when we do integration on polynomials.

By the power rule, we know that

$\int at^n dt= \frac{a}{n+1}t^{(n+1)} +C$ , where a, n, C are constants and is a variable.

$\int y''dt=\int (3t)dt$

$y'=\frac{3}{2}t^2+B$ , where is a constant

$\int y'dt=\int (\frac{3}{2}t^2+B)dt$

$y=\frac{1}{2}t^3+Bt+C$ , where and are constants. This is the most correct notation.

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Example Question #115 : How To Find Solutions To Differential Equations

In what interval(s) is the graph of the function $f(x)=x^2 + \frac{1}{x}$ concave down?

Possible Answers:

$(-\frac{\sqrt2}{2}, \frac{\sqrt2}{2})$

$(-\infty, 0)$

$(-\infty, 0) \cup (0, \infty)$

$(-\infty, -\frac{\sqrt2}{2}) \cup (\frac{\sqrt2}{2}, \infty)$

Never (The graph is always concave up.)

Correct answer:

$(-\frac{\sqrt2}{2}, \frac{\sqrt2}{2})$

Explanation:

The function is concave down when f''(x) < 0. $f'(x) = 2x+ \frac{1}{x} \Rightarrow f''(x) = 2-\frac{1}{x^2}.$

$f''(x)<0 \Rightarrow 2<\frac{1}{x^2} \Rightarrow 2x^2<1 \Rightarrow x^2< \frac{1}{2}. \\ \Rightarrow x \in (- \frac{\sqrt2}{2}, \frac{\sqrt2}{2}).$

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Example Question #2441 : Calculus

Find the general solution of the following differential equation:

$\frac{\mathrm{dy} }{\mathrm{d} x}=3y\sec^2(x)$

Possible Answers:

$y=Ce^{3\tan(x)}$

$y=Ce^{3\sec(x)tan(x)}$

y=x

$y=e^{3\tan(x)}+C$

Correct answer:

$y=Ce^{3\tan(x)}$

Explanation:

To find the general solution for the separable differential equation, we must move x and dx, y and dy to separate sides, and then integrate both sides:

$\int \frac{dy}{3y}=\int \sec^2(x)dx$

Next, integrate both sides:

$\frac{1}{3}\ln\left | y\right |=\tan(x)+C$

The rules used for the integrations are:

$\int \frac{dx}{x}=\ln\left | x\right |+C$ , $\int \sec^2(x)dx=\tan(x)+C$

Note that both Cs were combined to make one constant of integration in our equation.

Finally, solve for y:

$y=Ce^{3\tan(x)}$

Note that C was brought to the front, as e^C is itself a constant of integration.

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Calculus 1 : Equations

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #361 : Equations

Example Question #2441 : Calculus

Example Question #102 : How To Find Solutions To Differential Equations

Example Question #362 : Equations

Example Question #2444 : Calculus

Example Question #112 : How To Find Solutions To Differential Equations

Example Question #113 : How To Find Solutions To Differential Equations

Example Question #364 : Equations

Example Question #115 : How To Find Solutions To Differential Equations

Example Question #2441 : Calculus

All Calculus 1 Resources

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