There are 2 steps to solving this problem.

First, take the derivative of \(\displaystyle f(x)\).

Then, replace the value of x with the given point.

For example, if \(\displaystyle x=a\), then we are looking for the value of \(\displaystyle f'(x=a)\), or the derivative of \(\displaystyle f(x)\) at \(\displaystyle x=a\).

\(\displaystyle f(x)=4x^2-x+2\)

Calculate \(\displaystyle f'(x)\)

Derivative rules that will be needed here:

• Derivative of a constant is 0. For example, \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0\)
• Taking a derivative on a term, or using the power rule, can be done by doing the following: \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\)

\(\displaystyle f'(x)=8x-1\)

Then, plug in the value of x and evaluate

\(\displaystyle f'(x=1)=8*1-1 = 7\)

First we must find the first derivative of the function.

Because the derivative of the exponential function is the exponential function itelf, or

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}[e^x]=e^x\)

and taking the derivative is a linear operation,

we have that

\(\displaystyle f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}[f(x)]\)

\(\displaystyle =\frac{\mathrm{d} }{\mathrm{d} x}[2e^x]\)

\(\displaystyle =2\frac{\mathrm{d} }{\mathrm{d} x}[e^x]\)

\(\displaystyle =2e^x\)

Now setting \(\displaystyle x=0\)

\(\displaystyle f'(0)=2e^0=2*1=2\)

Thus

\(\displaystyle f'(0)=2\)

Find the rate of change of f(x) when x=3.

\(\displaystyle f(x)=3e^x-x^3+12x^2\)

To find a rate of change, we need to find the derivative.

First, recall the following rules:

\(\displaystyle 1) \frac{d}{dx}e^x=e^x\)

\(\displaystyle 2)\frac{d}{dx} x^n =nx^{n-1}\)

We can apply these two derivative rules to our function to get  our first derivative. Then we need to plug in 3 for x and solve.

\(\displaystyle f(x)=3e^x-x^3+12x^2\)

\(\displaystyle f'(x)=3e^x-3x^2+24x\)

\(\displaystyle f'(3)=3e^3-3(3)^2+24(3)=105.26\)

So, our answer is 105.26

To find \(\displaystyle g'(x)\), substitute \(\displaystyle u =\frac{ \pi }{x} = \pi x ^{-1}\) and use the chain rule: 

\(\displaystyle g(x) = \sec \left (\frac{\pi}{x} \right )\)

\(\displaystyle g'(x) = \frac{\mathrm{d} }{\mathrm{d} x} \sec\frac{\pi}{x}\)

\(\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} \sec u\)

\(\displaystyle = \frac{\mathrm{d} u}{\mathrm{d} x} \cdot \frac{\mathrm{d} }{\mathrm{d}u} \sec u\)

\(\displaystyle = \frac{\mathrm{d} }{\mathrm{d} x} \pi x^{-1} \cdot \frac{\mathrm{d} } {\mathrm{d}u} \sec u\)

\(\displaystyle = \pi \cdot (-1) \cdot x^{-1-1} \cdot \tan u \sec u\)

\(\displaystyle = - \pi x^{-2} \cdot \tan \frac{\pi}{x} \sec \frac{\pi}{x}\)

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

So  \(\displaystyle g '(x)= - \frac{\pi}{x^{2}}\tan \frac{\pi}{x} \sec \frac{\pi}{x}\)

and 

\(\displaystyle g '(6)= - \frac{\pi}{6^{2}}\tan \frac{\pi}{6} \sec \frac{\pi}{6}\)

\(\displaystyle = - \frac{\pi}{36} \cdot \frac{\sqrt{3}}{3} \cdot \frac{2\sqrt{3}}{3}\)

\(\displaystyle = - \frac{\pi}{36} \cdot \frac{2\cdot 3}{9} = - \frac{\pi}{54}\)

The slope of the line tangent to the graph of \(\displaystyle f(x)= 1+\frac{4}{x}\) at \(\displaystyle (2, 3)\) is

\(\displaystyle f'(2)\), which can be evaluated as follows:

\(\displaystyle f(x)= 1+\frac{4}{x} = 1 + 4x^{-1}\)

\(\displaystyle f'(x) = (-1) 4x^{-1-1}\)

\(\displaystyle f'(x) = - \frac{4}{x^{2}}\)

\(\displaystyle f'(2) = - \frac{4}{2^{2}}= - 1\)

The equation of the line with slope \(\displaystyle -1\) through \(\displaystyle (2, 3)\) is:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y - 3 = -1 (x-2)\)

\(\displaystyle y - 3 = - x+2\)

\(\displaystyle y = - x+5\)

The slope of the line tangent to the graph of \(\displaystyle f(x) = 2 \sin \frac{\pi x}{6}\) at \(\displaystyle (7, -1)\) is

\(\displaystyle f'(7)\), which can be evaluated as follows:

\(\displaystyle f(x) = 2 \sin \frac{\pi x}{6} = 2 \sin \frac{\pi }{6}x\)

\(\displaystyle f'(x) = 2 \cdot\frac {\pi }{6} \cos \frac{\pi }{6}x\)

\(\displaystyle f'(x) = \frac {\pi }{3} \cos \frac{\pi }{6}x\)

\(\displaystyle f'(7) = \frac {\pi }{3} \cos \frac{7 \pi }{6}\)

\(\displaystyle = \frac {\pi }{3} \left ( - \frac{\sqrt{3}}{2}\right ) = - \frac {\pi \sqrt{3 }}{6}\), the slope of the line.

The equation of the line with slope \(\displaystyle - \frac {\pi \sqrt{3 }}{6}\) through \(\displaystyle (7, -1)\) is:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y - (-1) = - \frac {\pi \sqrt{3 }}{6} (x-7)\)

\(\displaystyle y +1 = - \frac {\pi \sqrt{3 }}{6} x+\left ( \frac {\pi \sqrt{3 }}{6} \right ) 7\)

\(\displaystyle y = - \frac {\pi \sqrt{3 }}{6} x+ \frac {7 \pi \sqrt{3 }}{6} -1\)

The slope of the line tangent to the graph of \(\displaystyle f(x)= \frac{8}{x}\) at \(\displaystyle (-2,-4)\) is

\(\displaystyle f'(-2)\), which can be evaluated as follows:

\(\displaystyle f(x)= \frac{8}{x} = 8x^{-1}\)

\(\displaystyle f'(x)= 8 \cdot\left ( -1 x^{-2} \right )\)

\(\displaystyle f'(x)= -\frac{8}{x^{ 2}}\)

\(\displaystyle f'(-2)= -\frac{8}{(-2)^{ 2}} = -\frac{8}{4} = -2\), the slope of the line.

The equation of the line with slope \(\displaystyle -2\) through \(\displaystyle (-2,-4)\) is:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y - (-4) = -2 [x- (-2)]\)

\(\displaystyle y +4 = -2 (x+2)\)

\(\displaystyle y +4 = -2 x-4\)

\(\displaystyle y = -2 x-8\)

The slope of the line tangent to the graph of \(\displaystyle f(x)= \log (x+3)\) at the point \(\displaystyle (7,1)\) is \(\displaystyle f'(7)\), which can be evaluated as follows:

\(\displaystyle f(x)= \log (x+3)\)

\(\displaystyle f(x)=\frac{ \ln (x+3) }{\ln 10}\)

\(\displaystyle f'(x)=\frac{\mathrm{d} }{\mathrm{d} x} \frac{ \ln (x+3) }{\ln 10}\)

\(\displaystyle f'(x)= \frac{1 }{\ln 10} \cdot \frac{\mathrm{d} }{\mathrm{d} x}\ln (x+3)\)

\(\displaystyle f'(x)= \frac{1 }{\ln 10} \cdot \frac{1}{x+3}\)

\(\displaystyle f'(7)= \frac{1 }{\ln 10} \cdot \frac{1}{7+3}\)

\(\displaystyle f'(7)= \frac{1 }{10 \ln 10}\)

The line with this slope through \(\displaystyle (7,1)\) has equation:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y - 1= \frac{1 }{10 \ln 10} (x-7)\)

\(\displaystyle y - 1= \frac{1 }{10 \ln 10} \cdot x- \frac{1 }{10 \ln 10} \cdot 7\)\(\displaystyle y = \frac{x }{10 \ln 10} - \frac{7 }{10 \ln 10} +1\)

\(\displaystyle y = \frac{x -7 + 10 \ln 10}{10 \ln 10}\)

The slope of the line tangent to the graph of \(\displaystyle f(x) = x^{3} + x - 17\) at the point \(\displaystyle (3, 13)\) is \(\displaystyle f'(3)\), which can be evaluated as follows:

\(\displaystyle f(x) = x^{3} + x - 17\)

\(\displaystyle f'(x) = \frac{\mathrm{d} }{\mathrm{d} x}\left ( x^{3} + x - 17 \right )\)

\(\displaystyle f'(x) = 3x^{2} + 1\)

\(\displaystyle f'(3) = 3\cdot 3^{2} + 1 = 28\)

The line with slope 28 through \(\displaystyle (3, 13)\) has equation:

\(\displaystyle y - y_{1} = m (x-x_{1})\)

\(\displaystyle y -13= 28 (x-3)\)

\(\displaystyle y -13= 28 x- 28 \cdot 3\)

\(\displaystyle y -13= 28 x- 84\)

\(\displaystyle y = 28 x- 71\)

To find the slope at a point of a function, take the derivative of the function.

\(\displaystyle y=log_e(x)+5x\)

The derivative of \(\displaystyle log_a(x)\) is \(\displaystyle \frac{1}{x\:ln(a)}\).    

Therefore the derivative becomes,

\(\displaystyle \frac{1}{x\:ln(e)}+5=\frac{1}{x}+5\) since \(\displaystyle ln(e)=1\).

Now we substitute the given point to find the slope at that point.

\(\displaystyle \frac{dy}{dx}=\frac{1}{x} +5 = \frac{1}{10}+5 =\frac{51}{10}\)