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.

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

\(\displaystyle =5324+1089\)

\(\displaystyle =6413\)

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.

\(\displaystyle 17\)

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.

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

\(\displaystyle 125-15=110\)

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.

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

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,

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

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

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,

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

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

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

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

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

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

\(\displaystyle (866.9,305.4)\)

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

\(\displaystyle (16,0)\)

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

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