The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of a natural log: 

\(\displaystyle d[ln(u)]=\frac{du}{u}\)

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

Taking the derivative of the function \(\displaystyle f(x)=ln(x^6)\) at the point \(\displaystyle x=1\)

The slope of the tangent is

\(\displaystyle f'(x)=\frac{6x^5}{x^6}=\frac{6}{x}\)

\(\displaystyle f'(1)=\frac{6}{1}=6\)

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of a natural log: 

\(\displaystyle d[ln(u)]=\frac{du}{u}\)

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

Taking the derivative of the function

The slope of the tangent is \(\displaystyle f(x)=ln(2x^4)\) at the point \(\displaystyle x=2\)

\(\displaystyle f'(x)=\frac{8x^3}{2x^4}=\frac{4}{x}\)

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

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of a natural log: 

\(\displaystyle d[ln(u)]=\frac{du}{u}\)

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

Taking the derivative of the function \(\displaystyle f(x)=ln(x^2+x)\) at the point \(\displaystyle x=3\)

The slope of the tangent is

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

\(\displaystyle f'(3)=\frac{2(3)+1}{3^2+3}=\frac{7}{12}\)

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

\(\displaystyle d[e^u]=e^udu\)

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

Taking the derivative of the function \(\displaystyle f(x)=e^{x^3}\) at the point \(\displaystyle x=2\)

The slope of the tangent is

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

\(\displaystyle f'(2)=3(2)^2e^{2^3}=12e^8\)

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

\(\displaystyle d[e^u]=e^udu\)

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

Taking the derivative of the function \(\displaystyle f(x)=5e^{5x}\) at the point \(\displaystyle x=3\)

The slope of the tangent is

\(\displaystyle f'(x)=25e^{5x}\)

\(\displaystyle f'(3)=25e^{5(3)}=25e^{15}\)

The slope of the tangent can be found by taking the derivative of the function and evaluating the value of the derivative at a point of interest.

We'll need to make use of the following derivative rule(s):

Derivative of an exponential: 

\(\displaystyle d[e^u]=e^udu\)

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

Taking the derivative of the function \(\displaystyle f(x)=e^{2x+4}\) at the point \(\displaystyle x=4\)

The slope of the tangent is

\(\displaystyle f'(x)=2e^{2x+4}\)

\(\displaystyle f'(4)=2e^{2(4)+4}=2e^{12}\)

The first derivative of the given function makes use of the power rule of derivatives. 

If you have a function

\(\displaystyle f(x)=ax^n\)

According to the Power Rule, the first derivative is defined as 

\(\displaystyle f'(x)=nax^n^-^1\)

If this rule is applied to each of the terms in the function f(x) we get that 

\(\displaystyle f'(x)=4(2)x^4^-^1-3(3)x^3^-^1+2(2)x^2^-^1-1(18)x^1^-^1+0(5)x^0^-^1\)

\(\displaystyle =8x^3-9x^2+4x-18\)

To find the first derivative of this particular function is accomplished by applying the power rule which states,

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n= n(x)^{n-1}\)

Applying the above rule to the equation,

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

results in,

\(\displaystyle \\f'(x)= 2\cdot 3x^{2-1}+1\cdot 2x^{1-1} \\f'(x)=6x^1+2x^0\)

\(\displaystyle f'(x)= 6x+2\)

To take the derivative, you first use the power rule for differentiating, 

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n= nx^{n-1}\),

then you use the chain rule, 

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

Also recall the trigonometric rule for differentiating sine,

\(\displaystyle (sin(u))'=cos(u)\frac{du}{dx}\)

This produces,

\(\displaystyle f'(x)= 2\cdot 4x^{2-1} + \cos(2x)\cdot 1\cdot2x^{1-1}\)

\(\displaystyle f'(x)= 8x +2\cos(2x)\). 

Use the quotient rule to find the derivative which states,

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

Given,

\(\displaystyle f(x)=4x^2+3x\)

\(\displaystyle g(x)=x^3\)

the derivatives can be found using the power rule which states,

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

therefore,

\(\displaystyle \\ f'(x)=2\cdot 4x^{2-1}+1\cdot 3x^{1-1}=8x+3 \\g'(x)=3x^{3-1}=3x^2\)

Applying the quotient rule to our function we find the derivative to be as follows.

\(\displaystyle \frac{d}{dx}\frac{4x^2+3x}{x^3}=\frac{x^3(8x+3)-(4x^2+3x)3x^2}{x^6}\)

Simplify.

\(\displaystyle \frac{8x^4+3x^3-12x^4-9x^3}{x^6}\)

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

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