To find the derivative of the sum, we take the derivative of each term independently, then add them all up. Further, we use the rule

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

$\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(12x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(\sin(x))-\frac{\mathrm{d} }{\mathrm{d} x}(100)$

$\displaystyle f'=24x+\cos(x)$

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$, we solve

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(5x)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=4x+5$

To find the derivative of the function, we use the product rule, which is defined as

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(f*g)=(f'*g)+(f*g')$, where f and g are both functions.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(2x^5\sin{x})=\frac{\mathrm{d} }{\mathrm{d} x}(2x^5)*(\sin{x})+\frac{\mathrm{d} }{\mathrm{d} x}(\sin{x})*(2x^5)=10x^4\sin{x}+2x^5\cos{x}$

To find the derivative of the function using the quotient rule, we apply the following definition:

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{f}{g})=\frac{gf'-fg'}{g^2}$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(\frac{e^x}{\tan{x}})=\frac{\tan{x}*\frac{\mathrm{d} }{\mathrm{d} x}(e^x)-e^x*\frac{\mathrm{d} }{\mathrm{d} x}(\tan{x})}{\tan^2{x}}=\frac{e^x\tan{x}-e^x\sec^2{x}}{\tan^2{x}}$

To find the derivative of the function, we take the derivative of each element in the function independently, then add them up.

Using $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$, we solve

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(10x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(-1)=30x^2+10x$

Finding our second derivative requires two steps, we first must find the derivative then find the corresponding rate of change for that new equation.

$\displaystyle f(x)=ln(x^2-1)$

Here, the chain rule is used since our function is of the form $\displaystyle [f(g(x))]dx=f'(x)g'(f(x))$

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

We now must use the quotient rule since our function is a rational function. We use the rule $\displaystyle [\frac{f(x)}{g(x)}]dx=\frac{f'(x)g(x)-g'(x)f(x))}{(g(x))^2}$

Therefore,

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

To solve this problem, we can use the power rule. That means we lower the exponent of the variable by one and multiply the variable by that original exponent.

We're going to treat $\displaystyle 5$ as $\displaystyle 5x^0$, as anything to the zero power is one.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})+(0*5x^{0-1})$

Notice that $\displaystyle (0*5x^{0-1})=0$, as anything times zero is zero.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{3-1})+(1*2x^{1-1})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=(3*x^{2})+(2x^{0})$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}(x^3+2x+5)=3x^2+2$

First we need to find the derivative of the function. $\displaystyle f'(x)=\frac{1+x}{\sqrt{-3+2x+x^2}}$

Now, we can plug in $\displaystyle x=2$ to the derivative function.

$\displaystyle f'(2)=\frac{3\sqrt{5}}{5}$

To find the derivative of the function, we use a combination of the quotient rule and product rule

$\displaystyle f=\frac{x^4\sin{x}}{x^2+3}$

$\displaystyle f'=\frac{(x^2+3)(x^4\sin{x})'-x^4\sin{x}(x^2+3)'}{(x^2+3)^2}$

$\displaystyle f'=\frac{(x^2+3)(4x^3\sin{x}+x^4\cos{x})-x^4\sin{x}(2x)}{(x^2+3)^2}$

$\displaystyle f'=\frac{(x^2+3)(4x^3\sin(x)+x^4\cos(x)-2x^5\sin(x))}{(x^2+3)^2}$

To find the derivative of the function, we use the rule

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}x^n=nx^{n-1}$ and apply it to each term in the function

$\displaystyle f'=\frac{\mathrm{d} }{\mathrm{d} x}(x^3)+\frac{\mathrm{d} }{\mathrm{d} x}(5x^2)+\frac{\mathrm{d} }{\mathrm{d} x}(8x)+\frac{\mathrm{d} }{\mathrm{d} x}(20)$

$\displaystyle f'=3x^2+10x+8$