The rate of change of a function at a point is the value of the derivative at that point. First, take the derivative of f(x) using the power rule for each term.

Remember that the power rule is 

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\), and that the derivative of a constant is zero.

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

Next, notice that the x-value of the point (1,6) is 1, so substitute 1 for x in the derivative.

\(\displaystyle f'(1)=2+12(1)^3=14\)

Therefore, the rate of change of f(x) at the point (1,6) is 14. 

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

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

Need to use the power rule which states: \(\displaystyle \frac{d}{dx}u^n=nu^{n-1}\frac{du}{dx}\)

In our problem \(\displaystyle \frac{du}{dx}=-3sin(3x)\)

 \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}\), and the derivative of a constant is 0, so

\(\displaystyle f (x) = 7x^{2} + 6x - 9\)

\(\displaystyle f ' (x) = 7\cdot 2\cdot x^{2-1} + 6\cdot 1 + 0\)

\(\displaystyle f ' (x) = 14\cdot x^{1} + 6\)

\(\displaystyle f ' (x) = 14x + 6\)

First, find the derivative \(\displaystyle f'(x)\) of \(\displaystyle f(x)= 8x^{3}-7x^{2}+11\).

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}\), and the derivative of a constant is 0, so

\(\displaystyle f(x)= 8x^{3}-7x^{2}+11\)

\(\displaystyle f'(x)= 3 \cdot 8x^{3-1}-2\cdot 7x^{2-1}+0\)

\(\displaystyle f'(x)= 24x^{2}-14x^{1}\)

\(\displaystyle f'(x)= 24x^{2}-14x\)

Now, differentiate \(\displaystyle f'(x)\) to get \(\displaystyle f''(x)\).

\(\displaystyle f''(x)= 2 \cdot 24x^{2-1}-1\cdot 14x^{1-1}\)

\(\displaystyle f''(x)= 48x^{1}-14x^{0}\)

\(\displaystyle f''(x)= 48x-14\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (x^{n} \right )= n\cdot x^{n-1}\), so

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left ( x^{999}\right ) = 999x^{999-1} = 999x^{998}\)

Find the derivative of \(\displaystyle x^{777}\), then find the derivative of that expression.

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}\), so

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (x^{777} \right )= 777\cdot x^{777-1} = 777x^{776}\)

\(\displaystyle \frac{\mathrm{d^{2}} }{\mathrm{d} x^{2}} \left ( x^{777}\right ) = \frac{\mathrm{d} }{\mathrm{d} x} \left ( 777x^{776} \right )= 777\cdot776\cdot x^{776-1} = 602,952x^{775}\)

\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}\), and the derivative of a constant is 0, so

\(\displaystyle f(x)= 0.8x^{3}-0.5x^{2}+ 0.9\)

\(\displaystyle f'(x)= 3\cdot 0.8 x^{3-1}-2\cdot 0.5x^{2-1}+ 0\)

\(\displaystyle f'(x)= 2.4x^{2}-1x^{1}\)

\(\displaystyle f'(x) = 2.4x^{2}-x\)

First, find the derivative \(\displaystyle f'(x)\) of \(\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x\).

Recall that \(\displaystyle \frac{\mathrm{d} }{\mathrm{d} x} \left (ax^{n} \right )= n\cdot ax^{n-1}\), and the derivative of a constant is 0.

\(\displaystyle f(x) =0.9x^{4} + 0.7x^{3}- 0.5 x\)

\(\displaystyle f'(x) =4\cdot 0.9x^{4-1} + 3\cdot 0.7x^{3-1}- 1 \cdot 0.5 x ^{1-1}\)

\(\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5 x ^{0}\)

\(\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5\)

Now, differentiate \(\displaystyle f'(x)\) to get \(\displaystyle f''(x)\).

\(\displaystyle f'(x) =3.6x^{3} + 2.1x^{2}- 0.5\)

\(\displaystyle f''(x) =3 \cdot 3.6x^{3-1} + 2 \cdot 2.1x^{2-1}- 0\)

\(\displaystyle f''(x) =10.8x^{2} + 4.2x^{1}- 0\)

\(\displaystyle f''(x) =10.8x^{2} + 4.2x\)