The first term of the function's derivative is found simply with the power rule:

\(\displaystyle \frac{d}{dx}(x^n)=nx^{n-1}\rightarrow \frac{d}{dx}(4x^{-3})=-12x^{-4}\)

The second term is a trig derivative with the chain rule:

\(\displaystyle \frac{d}{dx}(6\sin(2x))=6\cos(2x)\cdot 2=12\cos(2x)\)

The third term requires the rule for derivatives of exponential and the chain rule:

\(\displaystyle \frac{d}{dx}(e^x)=e^x\rightarrow \frac{d}{dx}(3e^{4x})=3e^{4x}\cdot 4=12e^{4x}\)

Combining these three terms, we get the final answer:

\(\displaystyle 12e^{4x}+12\cos(2x)-12x^{-4} =12(e^{4x}+\cos(2x)-x^{-4})\)

The rule for the derivative of the product of two functions is as follows:

\(\displaystyle \frac{d}{dx}[f(x)\cdot g(x)]=f'(x)\cdot g(x)+f(x)\cdot g(x)\)

For the expression given, the two functions are:

\(\displaystyle f(x)=3x^2+7, \: \: \:g(x)=4x^3+x\)

Whose derivatives are:

\(\displaystyle f'(x)=6x,\: \: \:g'(x)=12x^2+1\)

The derivative of the product is then:

\(\displaystyle \frac{d}{dx}[f(x)\cdot g(x)]=(6x)(4x^3+x)+(3x^2+7)(12x^2+1)\)

Simplified, this reduces to:

\(\displaystyle 24x^4+6x^2+36x^4+3x^2+84x^2+7=60x^4+93x^2+7\)

The rule for the derivative of the product of two functions is as follows:

\(\displaystyle \frac{d}{dx}[f(x)\cdot g(x)]=f'(x)\cdot g(x)+f(x)\cdot g(x)\)

For the expression given, the two functions are:

\(\displaystyle f(x)=4\sin(2x), \: \: \:g(x)=e^{3x}\)

Whose derivatives are:

\(\displaystyle f'(x)=8\cos(2x),\: \: \:g'(x)=3e^{3x}\)

The derivative of the product is then:

\(\displaystyle \frac{d}{dx}[f(x)\cdot g(x)]=8\cos(2x)\cdot e^{3x}+4\sin(2x)\cdot 3e^{3x}\)

Simplifying, we get:

\(\displaystyle \frac{d}{dx}[f(x)\cdot g(x)]=8\cos(2x)\cdot e^{3x}+12\sin(2x)\cdot e^{3x} =4e^{3x}(2\cos(2x)+3\sin(2x))\)

Step 1: Use the power rule on the first term...

\(\displaystyle 2x^2\) becomes \(\displaystyle 4x\)

Step 2: Take derivative of second term...

\(\displaystyle 4\) becomes \(\displaystyle 0\)

Note: The derivative of a(n) constant term is always zero.

The first derivative is \(\displaystyle 4x\)

Step 1: Find the derivative of \(\displaystyle f(x)\) and \(\displaystyle g(x)\), which we denote as \(\displaystyle f'(x)\) and \(\displaystyle g'(x)\).

\(\displaystyle f'(x)=(1)'=0\)

\(\displaystyle g'(x)=(3x^2)'=6x\)

Step 2: Plug in the functions into the quotient rule formula:

\(\displaystyle \frac {0(3x^2)-6x(1)}{(3x^2)^2}\)

Step 3: Simplify...

\(\displaystyle \frac {0-6x}{9x^4}=\frac {-6x}{9x^4}\)

Step 4: Reduce:

\(\displaystyle \frac {-6x}{9x^4}=-\frac {2}{3x^3}\)

The first derivative is \(\displaystyle -\frac {2}{3x^3}\)

Step 1: We just take the derivative of \(\displaystyle f'(x)\), which is given to us in the question.

\(\displaystyle (2x^3-7x)'\)

Step 2: Use the power rule to take the derivative..

First Term: \(\displaystyle 2x^3\rightarrow 6x^2\)
Second Term: \(\displaystyle -7x\rightarrow-7\)

Step 3: Take the final answers in Step 2 and add them together...

The second derivative, \(\displaystyle f''(x)\), is \(\displaystyle 6x^2-7\)

Recall that when taking the derivative, multiply the exponent by the coefficient in front of the x term and then subtract one from the exponent:

\(\displaystyle f{}'(x)=9x^2+x\)

Step 1: Re-write \(\displaystyle \large f(x)\) as \(\displaystyle \large x\) to a power:

\(\displaystyle \large \sqrt{x}=x^\frac {1}{2}\)
Step 2: Use the power rule:

\(\displaystyle \large x^\frac {1}{2}=\frac {1}{2}x^{\frac {1}{2}-1}\)

Step 3: Simplify the exponent:

\(\displaystyle \large \frac {1}{2}x^{\frac {1}{2}}-1=\frac {1}{2}x^\frac {-1}{2}\)

Step 4: Re-write \(\displaystyle \large x^\frac {-1}{2}\) as a fraction with a positive exponent:

\(\displaystyle \large x^\frac {-1}{2}=\frac {1}{x^\frac {1}{2}}=\frac {1}{\sqrt{x}}\)
Step 5: Multiply the result in Step 4 with the coefficient \(\displaystyle \large \frac {1}{2}\) in step 3:

\(\displaystyle \large \frac {1}{2} \cdot \frac {1}{\sqrt{x}}=\frac {1}{2\sqrt{x}}\)

\(\displaystyle \begin{align*}&\text{The function }f(x)=-11cos(14x^{2})\text{, can be seen as a function nested within another.}\\&\text{Notice the }14x^{2}\text{ within the }-11cos()\\&\text{We'll wish to use the chain rule:}\\&(f\circ g )'=(f'\circ g )\cdot g'\\&\text{Along with the following:}\\&d[cos(u)]=-sin(u)du\\&u=14x^{2}\\&du=28x\\&f'(x)=308xsin(14x^{2})\end{align*}\)