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Calculus 1 : How to find differential functions

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Example Questions

Example Question #1472 : Calculus

Derive the following function.

$\sin(3x^2+7)$

Possible Answers:

$\cos(6x)$

$\left 6x[\sin(3x^2+7)]$

$\left 6x[\cos(3x^2+7)]$

$\left 3x^2[\cos(3x^2+7)]$

$\left -6x[\cos(3x^2+7)]$

Correct answer:

$\left 6x[\cos(3x^2+7)]$

Explanation:

By deriving the trigonometric function $\sin(x)$ , we are given $x'\cos(x)$ . The is usually ignored unless is being used to express a different function, where it's own derivative is not . In the case of $\sin(3x^2+7)$ , we take the derivative of not only the $\sin$ function but also the 3x^2+7 . We then multiply the two derivatives and end up with our answer, $\left 6x[\cos(3^2+7)]$ .

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Example Question #1473 : Calculus

What is the slope of the function $f(xy)=\frac{1}{e^{xy}}$ at the point (-1,3) ?

Possible Answers:

2e^3

$-3e^3\widehat{i}+e^3\widehat{j}$

$3e^3\widehat{i}-e^3\widehat{j}$

$-e^3\widehat{i}+e^3\widehat{j}$

-2e^3

Correct answer:

$-3e^3\widehat{i}+e^3\widehat{j}$

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[e^u]=e^udu

Note that u and v 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.

Evaluating the derivatives of $f(xy)=\frac{1}{e^{xy}}=e^{-xy}$ at the point (1,3)

$\frac{\delta f}{\delta x}=-ye^{-xy};\frac{\delta f}{\delta x}=-3e^{-(-1)3}=-3e^3$

$\frac{\delta f}{\delta y}=-xe^{-xy};\frac{\delta f}{\delta y}=-(-1)e^{-(-1)(3)}=e^3$

Thus the slope is

$-3e^3\widehat{i}+e^3\widehat{j}$

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Example Question #1474 : Calculus

What is the slope of the function $f(x,y)=x^2y+\frac{x}{y^2}$ at the point (3,1) ?

Possible Answers:

$11\widehat{i}+23\widehat{j}$

$2\widehat{i}+5\widehat{j}$

$7\widehat{i}+3\widehat{j}$

$3\widehat{i}+1\widehat{j}$

$-4\widehat{i}-1\widehat{j}$

Correct answer:

$7\widehat{i}+3\widehat{j}$

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

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.

Taking the partials of $f(x,y)=x^2y+\frac{x}{y^2}=x^2y+xy^{-2}$ at the point (3,1)

$\frac{\delta f}{\delta x}=2xy+\frac{1}{y^2}=2(3)(1)+\frac{1}{1^2}=7$

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

Thus the slope is $7\widehat{i}+3\widehat{j}$

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Example Question #264 : Other Differential Functions

What is the slope of the function $f(x,y,z) = \frac{x}{y+z}$ at the point (-1,0,2) ?

Possible Answers:

$0.5\widehat{i}-0.25\widehat{j}-0.25\widehat{k}$

$-0.25\widehat{i}-0.25\widehat{j}+0.25\widehat{k}$

$0.5\widehat{i}+0.25\widehat{j}+0.25\widehat{k}$

$0.25\widehat{i}+0.25\widehat{j}+0.25\widehat{k}$

$1\widehat{i}+0.5\widehat{j}+0.5\widehat{k}$

Correct answer:

$0.5\widehat{i}+0.25\widehat{j}+0.25\widehat{k}$

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.

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.

Taking the partials of $f(x,y,z) = \frac{x}{y+z}=x(y+z)^{-1}$ at the point (-1,0,2)

$\frac{\delta f}{\delta x}=\frac{1}{y+z}=\frac{1}{0+2}=0.5$

$\frac{\delta f}{\delta y}=\frac{-x}{(y+z)^2}=\frac{-(-1)}{(0+2)^2}=0.25$

$\frac{\delta f}{\delta z}=\frac{-x}{(y+z)^2}=\frac{-(-1)}{(0+2)^2}=0.25$

The slope is

$0.5\widehat{i}+0.25\widehat{j}+0.25\widehat{k}$

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Example Question #265 : Other Differential Functions

Bored, yet staggeringly intelligent cave men are catapulting fruit from some distance to the left into a bowl-shaped area. Ten feet to the right of the base of this "bowl" is an entrance to a competing tribe's cave which is a linear, angled tunnel orthogonal to the bowl (that is, perpendicular to the tangent line at the surface). The x-axis length of the tube is $5 \:\uptext{ft}$ . The bowl's shape is given by:

$f(x)= \frac{1}{32}x^{2}$ , where negative values on the x-axis give how far to the left

someone is from base (or origin) of the bowl.

At what elevation does a piece of fruit enter the cave from the bottom of the tunnel?

Possible Answers:

$-\frac{15}{64}$

$-\frac{39}{8}$

$-\frac{15}{64}$

$\frac{65}{32}$

Correct answer:

$-\frac{39}{8}$

Explanation:

We want to find the equation of the normal line to f(x) at x=10 . The reason for this is that if the fruit goes down a straight tunnel at a right angle to the tangent at that point, its trajectory is uniquely given by the perpendicular line at that point. That is the very definition of a normal line. The slope of the tangent line, of course, is given by the derivative:

$f(x)= \frac{1}{32}x^{2}$

$f'(x)= \frac{2}{32}x$ so...

$f'(10)= \frac{20}{32} = \frac{5}{8}$

We know the slope of a perpendicular line has negative reciprocal slope of the tangent, so the slope of the normal line is :

$\frac{-1}{\frac{5}{8}}= -\frac{8}{5}$

So, using the point slope formula, the formula for the normal line is:

$y- \frac{25}{8} = -\frac{8}{5}(x-10)$

$y= 16-\frac{8}{5}x +\frac{25}{8} = \frac{153}{8}-\frac{8}{5}x$

Now we use this to find the end of the tunnel. Since the x-coordinate of the tunnels length is five, we know its exit is at x=15 , so the fruit exits the tunnel at:

$\frac{153}{8}-\frac{8}{5}(15) = -\frac{39}{8}$

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Example Question #266 : Other Differential Functions

Find the slope of the function f(x,y)=x^2ycos(y) at the point $(5,2\pi)$

Possible Answers:

$4\widehat{i}+5\pi\widehat{j}$

$4\pi\widehat{i}+5\widehat{j}$

$20\pi\widehat{i}+25\widehat{j}$

$20\widehat{i}+25\widehat{j}$

$4\widehat{i}+5\widehat{j}$

Correct answer:

$20\pi\widehat{i}+25\widehat{j}$

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 rules will be necessary:

Trigonometric derivative: d[cos(u)]=-sin(u)du

Product rule: d[uv]=udv+vdu

Note that u and v 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.

Taking the partials of f(x,y)=x^2ycos(y) at the point $(5,2\pi)$

$\frac{\delta f}{\delta x}=2xycos(y)=2(5)(2\pi)cos(2\pi)=20\pi$

$\frac{\delta f}{\delta y}=x^2cos(y)-x^2ysin(y)=5^2cos(2\pi)-5^2(2\pi)sin(2\pi)=25$

The slope is

$20\pi\widehat{i}+25\widehat{j}$

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Example Question #267 : Other Differential Functions

Find the slope of the function f(x,y)=ysin(2x)+2xsin(y) at the point $(\pi,\pi)$

Possible Answers:

$2\pi\widehat{i}-2\pi\widehat{j}$

$\pi\widehat{i}+\pi\widehat{j}$

$2\pi\widehat{i}+2\pi\widehat{j}$

$2\widehat{i}+2\widehat{j}$

$2\widehat{i}-2\widehat{j}$

Correct answer:

$2\pi\widehat{i}-2\pi\widehat{j}$

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 and v 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.

Taking the derivatives of f(x,y)=ysin(2x)+2xsin(y) at the point $(\pi,\pi)$

$\frac{\delta f}{\delta x}=2ycos(2x)+2sin(y)=2\pi cos(2\pi)+2sin(\pi)=2\pi$

$\frac{\delta f}{\delta y}=sin(2x)+2xcos(y)=sin(2\pi)+2\pi cos(\pi)=-2\pi$

The slope is

$2\pi\widehat{i}-2\pi\widehat{j}$

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Example Question #261 : How To Find Differential Functions

Find the slope of the function f(x,y)=e^xln(y)sin(xy) at the point (3,3)

Possible Answers:

$11.1\widehat{i}-0.8 \widehat{j}$

$-51.2\widehat{i}-57.6\widehat{j}$

$12.3\widehat{i}+81.2 \widehat{j}$

$63.9\widehat{i}-112.8 \widehat{j}$

$-18.4\widehat{i}+23.9 \widehat{j}$

Correct answer:

$-51.2\widehat{i}-57.6\widehat{j}$

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 rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Derivative of a natural log: $d[ln(u)]=\frac{du}{u}$

Trigonometric derivative: d[sin(u)]=cos(u)du

Product rule: d[uv]=udv+vdu

Note that u and v 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.

Taking the partial derivatives of f(x,y)=e^xln(y)sin(xy) at the point (3,3)

$\frac{\delta f}{\delta x}=e^xln(y)sin(xy)+ye^xln(y)cos(xy)$

$\frac{\delta f}{\delta x}=e^3ln(3)sin(3(3))+(3)e^3ln(3)cos(3(3))=-51.2$

$\frac{\delta f}{\delta y}=\frac{e^xsin(xy)}{y}+xe^xln(y)cos(xy)$

$\frac{\delta f}{\delta y}=\frac{e^3sin(3(3))}{3}+3e^3ln(3)cos(3(3))=-57.6$

The slope is

$-51.2\widehat{i}-57.6\widehat{j}$

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Example Question #262 : How To Find Differential Functions

Find the slope of the function $f(x,y,z)=x^2e^z+ye^{3z}$ at the point (3,4, ln(2))

Possible Answers:

$9\widehat{i}+32\widehat{j}+108\widehat{k}$

$25\widehat{i}+10\widehat{j}+6\widehat{k}$

$80\widehat{i}+15\widehat{j}+22\widehat{k}$

$44\widehat{i}+10\widehat{j}+56\widehat{k}$

$12\widehat{i}+8\widehat{j}+114\widehat{k}$

Correct answer:

$12\widehat{i}+8\widehat{j}+114\widehat{k}$

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[e^u]=e^udu

Note that u and v 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 $f(x,y,z)=x^2e^z+ye^{3z}$ at the point (3,4, ln(2))

$\frac{\delta f}{\delta x}=2xe^z=2(3)e^{ln(2)}=12$

$\frac{\delta f}{\delta y}=e^{3z}=e^{3ln(2)}=8$

$\frac{\delta f}{\delta z}=x^2e^z+3ye^{3z}=3^2e^{ln(2)}+3(4)e^{3ln(2)}=18+96=114$

The slope is

$12\widehat{i}+8\widehat{j}+114\widehat{k}$

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Example Question #270 : Other Differential Functions

What is the slope of the function f(x,y)=sin(xe^y) at the point (2,3) ?

Possible Answers:

$-33.4\widehat{i}-66.8\widehat{j}$

$56.4\widehat{i}+28.2\widehat{j}$

$11.8\widehat{i}+22.3\widehat{j}$

$-67.2\widehat{i}-11.6\widehat{j}$

$-15.7\widehat{i}-31.5\widehat{j}$

Correct answer:

$-15.7\widehat{i}-31.5\widehat{j}$

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 rules will be necessary:

Derivative of an exponential: d[e^u]=e^udu

Trigonometric derivative: d[sin(u)]=cos(u)du

Note that u and v 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(xe^y) at the point (2,3)

$\frac{\delta f}{\delta x}=e^ycos(xe^y)=e^3cos(2e^3)=-15.7$

$\frac{\delta f}{\delta y}=xe^ycos(xe^y)=2e^3cos(2e^3)=-31.5$

The slope is $-15.7\widehat{i}-31.5\widehat{j}$

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Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1472 : Calculus

Example Question #1473 : Calculus

Example Question #1474 : Calculus

Example Question #264 : Other Differential Functions

Example Question #265 : Other Differential Functions

Example Question #266 : Other Differential Functions

Example Question #267 : Other Differential Functions

Example Question #261 : How To Find Differential Functions

Example Question #262 : How To Find Differential Functions

Example Question #270 : Other Differential Functions

All Calculus 1 Resources

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