Given the function \(\displaystyle f(x)=10x^{2}-9x+1\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^{n}=nx^{n-1}\) for all \(\displaystyle n\neq0\) to find its derivative:

\(\displaystyle f'(x)=(2)10x^{2-1}-(1)9x^{1-1}+(0)=20x-9\).

Plugging in the \(\displaystyle x\)-value of the point \(\displaystyle (-2,7)\) into \(\displaystyle f'(x)\), we get 

\(\displaystyle f'(-2)=20(-2)-9=-40-9=-49\).

Use either the FOIL method to simplify before taking the derivative or use the product rule to find the derivative of the function.

The product rule will be used for simplicity.

\(\displaystyle y'=(2x+4)(3)+(3x+2)(2) = 6x+12 + 6x+4= 12x+16\)

Substitute \(\displaystyle x=1\).

\(\displaystyle y'(1)= 12(1)+16=28\)

The derivative of the function is given by the product rule:

\(\displaystyle h(x)=f(x)g(x)\), \(\displaystyle h'(x)=f'(x)g(x)+f(x)g'(x)\)

Simply find the derivative of each function:

\(\displaystyle \frac{d}{dx}(2x)=2\)

\(\displaystyle \frac{d}{dx}\sin(x)=\cos(x)\)

The derivatives were found using the following rules:

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

Simply evaluate each derivative and the original functions at the point given, using the above product rule.

Slope is defined as the first derivative of a given function.

Since \(\displaystyle f(x)=-2x^2+5x-3\), we can use the Power Rule

\(\displaystyle \frac{d}{dx}x^n=nx^{n-1}\) for all \(\displaystyle n\neq 0\) to determine that 

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

Since we're given a point \(\displaystyle (0,1)\), we can use the x-coordinate \(\displaystyle 0\) to solve for the slope at that point.

Thus, 

\(\displaystyle f'(0)=-4(0)+5=5.\)

The slope of the tangent line to a function at a point is the value of the derivative at that point. To calculate the derivative in this problem, the product rule is necessary. Recall that the product rule states that:

\(\displaystyle \frac{d}{dx}(m(x)n(x))=m'(x)n(x)+m(x)n'(x)\).

In this example, \(\displaystyle m(x)=e^x \ and\ n(x)=ln(x).\)

Therefore, 

\(\displaystyle m'(x)=e^x \ and\ n'(x)=\frac{1}{x}\), and

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

At x = 1, this dervative has the value

\(\displaystyle e^1ln(1)+e^1\cdot\frac{1}{1}=0+e=e\).

In order to find the derivative, we need to find \(\displaystyle f'(x)\). We can find this by remembering the product rule and knowing the derivative of natural log.

Product Rule:

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

Derivative of natural log:

\(\displaystyle f(x)=\ln(g(x))\)

\(\displaystyle f'(x)=\frac{g'(x)}{g(x)}\)

Now lets apply this to our problem.

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

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

\(\displaystyle f'(0)=3(0)^2\ln(0+1)+\frac{0^3}{0+1}=0\)

To find the second derivative of the function, we first must find the first derivative of the function:

\(\displaystyle f'(x)=2e^{2x}+2\cos(3x)-6x\sin(3x)\)

The derivative was found using the following rules:

\(\displaystyle \frac{d}{dx}\cos(x)=-\sin(x)\), \(\displaystyle \frac{d}{dx}e^u=e^u\frac{du}{dx}\), \(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\), \(\displaystyle \frac{d}{dx}f(g(x))=f'(g(x))\cdot g'(x)\), \(\displaystyle \frac{d}{dx}f(x)g(x)=f'(x)g(x)+f(x)g'(x)\)

The second derivative is simply the derivative of the first derivative function, and is equal to:

\(\displaystyle f''(x)=4e^{2x}-12\sin(3x)-18x\cos(3x)\)

One more rule used in combination with some of the ones above is:

\(\displaystyle \frac{d}{dx}\sin(x)=\cos(x)\)

To finish the problem, plug in x=0 into the above function to get an answer of \(\displaystyle 4\).

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 and evaluate

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 \(\displaystyle x=a\).

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

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:\(\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}\)

\(\displaystyle f'(x)=x^2 - 4x + 1\)

Then, plug in the value of x and evaluate

\(\displaystyle f'(x=3)=3^2-4*3+1=-2\)

The equation for the derivative at a point is given by

\(\displaystyle f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}\).

By substituting \(\displaystyle a = e+1\), \(\displaystyle f = \ln(2x)\), we obtain

\(\displaystyle f'(e+1)=\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}.\)