Calculus 1 : How to find differential functions

Study concepts, example questions & explanations for Calculus 1

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Example Questions

Example Question #1491 : Calculus

Find the slope of the tangent line at \(\displaystyle x=11\).

\(\displaystyle x^{3}(x+3)\)

Possible Answers:

\(\displaystyle 1893\)

\(\displaystyle 1794\)

\(\displaystyle 1894\)

\(\displaystyle 6413\)

Correct answer:

\(\displaystyle 6413\)

Explanation:

Begin by finding the derivative using the product rule.

The product rule is,

\(\displaystyle f(x)g(x)'=f(x)g'(x)+g(x)f'(x)\).

Applying this rule to our function where

\(\displaystyle f(x)=x^3\rightarrow f'(x)=3x^2\)

\(\displaystyle g(x)=x+3\rightarrow g'(x)=1\)

we get,

\(\displaystyle x^3(1)+(x+3)3x^2\)

\(\displaystyle x^3+3x^3+9x^2\)

\(\displaystyle 4x^3+9x^2\)

Now, substitute 11 for x.

\(\displaystyle 4(11)^{3}+9\cdot11^{2}\)

\(\displaystyle =5324+1089\)

\(\displaystyle =6413\)

Example Question #461 : Functions

Find the slope of the tangent line at \(\displaystyle x=1\).

\(\displaystyle (x+3)x^{4}\)

Possible Answers:

\(\displaystyle 21\)

\(\displaystyle 23\)

\(\displaystyle 19\)

\(\displaystyle 17\)

Correct answer:

\(\displaystyle 17\)

Explanation:

Begin by finding the derivative using the product rule.

The product rule is,

\(\displaystyle f(x)g(x)'=f(x)g'(x)+g(x)f'(x)\).

Applying this rule to our function where

\(\displaystyle f(x)=(x+3)\rightarrow f'(x)=1\)

\(\displaystyle g(x)=x^4\rightarrow g'(x)=4x^3\)

we get,

\(\displaystyle (x+3)4x^3+x^{4}(1)\)

\(\displaystyle 4x^{4}+12x^{3}+x^{4}\)

\(\displaystyle 5x^{4}+12x^{3}\)

Now, substitute 1 for x.

\(\displaystyle 17\)

Example Question #461 : Functions

What is the slope of the tanget line at \(\displaystyle x=2.5\)?

\(\displaystyle 2x^{4}-3x^{2}+2\)

Possible Answers:

\(\displaystyle 110\)

\(\displaystyle 115\)

\(\displaystyle 105\)

\(\displaystyle 100\)

Correct answer:

\(\displaystyle 110\)

Explanation:

First, find the derivative using the power rule which states, \(\displaystyle (x^n)'=nx^{n-1}\).

Applying the power rule to each term in the function we get,

\(\displaystyle 8x^{3}-6x\)

Now, plug in 2.5 for x.

\(\displaystyle 8(2.5^{3})-6(2.5)\)

\(\displaystyle 125-15=110\)

Example Question #281 : Other Differential Functions

Find the slope of the tangent line at \(\displaystyle x=1.5\).

\(\displaystyle \frac{x^{2}}{4}\)

Possible Answers:

\(\displaystyle 1\)

\(\displaystyle \frac{1}{2}\)

\(\displaystyle \frac{3}{4}\)

\(\displaystyle \frac{1}{4}\)

Correct answer:

\(\displaystyle \frac{3}{4}\)

Explanation:

First, find the derivative using the quotient rule which states,

\(\displaystyle \frac{f(x)}{g(x)}'=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\).

In this particular case,

\(\displaystyle f(x)=x^2\rightarrow f'(x)=2x\)

\(\displaystyle g(x)=4\rightarrow g'(x)=0\)

Therefore the derivative becomes,

\(\displaystyle \frac{4(2x)-x^2(0)}{4^2}=\frac{4(2x)}{16}\)

\(\displaystyle =\frac{8x}{16}=\frac{x}{2}\)

Now, substitute 1.5 for x.

\(\displaystyle \frac{1.5}{2}=\frac{3}{4}\)

Example Question #281 : Other Differential Functions

Find the derivative.

\(\displaystyle \frac{4}{2x}\)

Possible Answers:

\(\displaystyle -\frac{2}{x^{2}}\)

\(\displaystyle -x^{2}\)

\(\displaystyle -\frac{2}{x^{3}}\)

\(\displaystyle x\)

Correct answer:

\(\displaystyle -\frac{2}{x^{2}}\)

Explanation:

Use the quotient rule to find the derivative which states,

\(\displaystyle \frac{f(x)}{g(x)}'=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\).

In this particular case,

\(\displaystyle f(x)=4\rightarrow f'(x)=0\)

\(\displaystyle g(x)=2x\rightarrow g'(x)=2\).

Therefore the derivative becomes,

\(\displaystyle \frac{2x(0)-4(2)}{(2x)^2}=\frac{-8}{4x^{2}}\)

\(\displaystyle =-\frac{2}{x^{2}}\)

 

Example Question #1501 : Calculus

Find the derivative.

\(\displaystyle \frac{2x}{x^{3}}\)

Possible Answers:

\(\displaystyle \frac{-4}{x^{4}}\)

\(\displaystyle \frac{-4}{x^3}\)

\(\displaystyle \frac{-3}{x^{3}}\)

\(\displaystyle \frac{-2}{x^5}\)

Correct answer:

\(\displaystyle \frac{-4}{x^3}\)

Explanation:

Use the quotient rule to find the derivative which states,

\(\displaystyle \frac{f(x)}{g(x)}'=\frac{g(x)f'(x)-f(x)g'(x)}{g(x)^2}\).

In this particular case,

\(\displaystyle f(x)=2x\rightarrow f'(x)=2\)

\(\displaystyle g(x)=x^3\rightarrow g'(x)=3x^2\).

Therefore, the derivative becomes,

\(\displaystyle =\frac{2x^{3}-2x(3x^{2})}{x^{6}}\)

\(\displaystyle =\frac{-4x^{3}}{x^6}\)

\(\displaystyle =\frac{-4}{x^3}\)

Example Question #287 : Other Differential Functions

What is the slope of the function \(\displaystyle f(x,y)=x^y\) at \(\displaystyle (e,4)\)?

Possible Answers:

\(\displaystyle (4e^4,e^4\ln(4))\)

\(\displaystyle (4e^3,e^4)\)

\(\displaystyle (4e^3,e^4\ln(4))\)

\(\displaystyle (e^3,e^4)\)

\(\displaystyle (e^4,e^4)\)

Correct answer:

\(\displaystyle (4e^3,e^4)\)

Explanation:

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: \(\displaystyle d[a^u]=a^u\ln(a)du\)

Note that \(\displaystyle u\) 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 \(\displaystyle f(x,y)=x^y\) at \(\displaystyle (e,4)\)

\(\displaystyle x\):

\(\displaystyle \frac{\delta f}{\delta x}=yx^{y-1}=4e^3\)

\(\displaystyle y\):

\(\displaystyle \frac{\delta f}{\delta y}=x^y\ln(x)=e^4\ln(e)=e^4\)

The slope is

\(\displaystyle (4e^3,e^4)\)

Example Question #288 : Other Differential Functions

Find the slope of the function \(\displaystyle f(x,y)=x^yy^x\) at the point \(\displaystyle (2,4)\)

Possible Answers:

\(\displaystyle (305.4,866.9)\)

\(\displaystyle (457.2,203.3)\)

\(\displaystyle (203.3,457.2)\)

\(\displaystyle (866.9,305.4)\)

Correct answer:

\(\displaystyle (866.9,305.4)\)

Explanation:

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: \(\displaystyle d[a^u]=a^u\ln(a)du\)

Product rule: \(\displaystyle d[uv]=udv+vdu\)

Note that \(\displaystyle u\) and \(\displaystyle v\) 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.

Find the partial derivatives of \(\displaystyle f(x,y)=x^yy^x\) at the point \(\displaystyle (2,4)\)

\(\displaystyle x\):

\(\displaystyle \frac{\delta f}{\delta x}=x^{y-1}y^{x+1}+x^{y}y^{x}ln(y)=2^34^3+2^44^2ln(4)=866.9\)

\(\displaystyle y\):

\(\displaystyle \frac{\delta f}{\delta y}=x^{y}y^{x}ln(x)+x^{y+1}y^{x-1}=2^44^2ln(2)+2^54=305.4\)

The slope is

\(\displaystyle (866.9,305.4)\)

Example Question #289 : Other Differential Functions

Find the slope of the function \(\displaystyle f(x,y)=x^{2+2y}\) at \(\displaystyle (1,7)\).

Possible Answers:

\(\displaystyle (2,2)\)

\(\displaystyle (0,16)\)

\(\displaystyle (4,8)\)

\(\displaystyle (8,4)\)

\(\displaystyle (16,0)\)

Correct answer:

\(\displaystyle (16,0)\)

Explanation:

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: \(\displaystyle d[a^u]=a^u\ln(a)du\)

Note that \(\displaystyle u\) 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  \(\displaystyle f(x,y)=x^{2+2y}\) at \(\displaystyle (1,7)\)

\(\displaystyle x\):

\(\displaystyle \frac{\delta f}{\delta x}=(2+2y)x^{1+2y}=(2+14)1^{15}=16\)

\(\displaystyle y\):

\(\displaystyle \frac{\delta f}{\delta y}=2x^{2+2y}\ln(x)=2(1^{16})\ln(1)=0\)

The slope is

\(\displaystyle (16,0)\)

Example Question #290 : Other Differential Functions

Find the slope of the function \(\displaystyle f(x,y,z)=x^{yz^2}\) at the point \(\displaystyle (2,3,1)\)

Possible Answers:

\(\displaystyle (5.5,33.3,12)\)

\(\displaystyle (12,5.5,33.3)\)

\(\displaystyle (33.3,5.5,12)\)

\(\displaystyle (12,33.3,5.5)\)

\(\displaystyle (5.5,12,33.3)\)

Correct answer:

\(\displaystyle (12,5.5,33.3)\)

Explanation:

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

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

\(\displaystyle \frac{\delta f}{\delta x_1}\widehat{e_1}+\frac{\delta f}{\delta x_2}\widehat{e_2}+...+\frac{\delta f}{\delta x_n}\widehat{e_n}\)

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: \(\displaystyle d[a^u]=a^u\ln(a)du\)

Note that \(\displaystyle u\) 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 derivative of  \(\displaystyle f(x,y,z)=x^{yz^2}\) at the point \(\displaystyle (2,3,1)\)

\(\displaystyle x\):

\(\displaystyle \frac{\delta f}{\delta x}=yz^2x^{yz^2-1}=3(1^2)2^{3(1^2)-1}=12\)

\(\displaystyle y\):

\(\displaystyle \frac{\delta f}{\delta y}=z^2x^{yz^2}\ln(x)=1^2(2^{3(1^2)})\ln(2)=5.5\)

\(\displaystyle z\):

\(\displaystyle \frac{\delta f}{\delta z}=2yzx^{yz^2}\ln(x)=2(3)(1)(2^{3(1^2)})\ln(2)=33.3\)

The slope is

\(\displaystyle (12,5.5,33.3)\)

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