Use the power rule to find the derivative of the function.

In mathematical terms, the power rule states,

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

Move the exponent to the front, making it the coefficient.

Next, decrease the exponent by $\displaystyle 1$ making it $\displaystyle x^{\frac{-2}{3}}$.

After simplifying, you get 

$\displaystyle \frac{1}{3\sqrt[3]{x^2}}$.

To solve this problem, we need to use the identity that tells us that 

$\displaystyle \frac{d}{dx}(\sec (x))=\sec(x)\tan(x)$.

After using this identity, we tack the 3 back on to get 

$\displaystyle h{}'(x)=3\sec(x)\tan(x)$.

To solve this equation we will use power rule.

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. We will do this for all values in this problem that have an "x" value attached to it.

$\displaystyle x^3 = 3x^2 ; 3x^2 = 6x ; 3x=3$

The 3 on the end has no "x" so the derivative we will set that equal to just zero. Then combine like terms and express as a single derivative.

To find the derivative we're going to need to know trig and power rule derivative rules.

Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $\displaystyle (nx^{^{n-1}})$

And for trig, the derivative of $\displaystyle sinx=cosx$.

Therefore using these rules we get: $\displaystyle y'= cosx + 2x$

Quotient Rule states that we take $\displaystyle \frac{f(x)g'(x)-g(x)f'(x)}{(g(x))^2}$

$\displaystyle f(x)= x+1$ ; $\displaystyle f'(x)=1$

$\displaystyle g(x)=x-2$ ; $\displaystyle g'(x)=1$

Plug this into our formula and we get

$\displaystyle \frac{(x-2)-(x+1)}{(x-2)^2}$

To find the derivative we need to use chain rule.

Chain Rules states that we work from the outside to the inside. Meaning we will take the derivative of the outside of the equation and multiply it by the derivative of the inside of the equation.

$\displaystyle y= (x+2)^2$ ;The derivative of the outside will be  $\displaystyle 2(x+2)^1$

And the inside $\displaystyle (x+2)$ derivative will just be 1.

Multiply them together and we are left with

$\displaystyle 2(x+2)$

To find the derivative we need to use Product Rule and Power Rule. Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. To find the derivative we need to use product rule. Product rule states that we take the derivative of the first function and multiply it by the derivative of the second function and then add that with the derivative of the second function multiplied by the given first function. 

In equation this looks like:

$\displaystyle f(x)g'(x)+g(x)f'(x)$

And power rule looks like: $\displaystyle n(x^{n-1})$

We need to use power rule to find the derivatives of each individual function (g(x) and f(x))

$\displaystyle f(x)=x^2 ; f'(x) = 2x$

$\displaystyle g(x)= (x^2+3) ; g'(x)= 2x$

Fortunately for this problem the derivatives are the same for each function so let's plug this into the Rule

$\displaystyle x^2(2x) + (x^2 +3)(2x)$

$\displaystyle = 4x^3 + 2x$

To find the derivative we need to use power rule. Power rule says that we take the exponent of the “x” value and bring it to the front. Then we subtract one from the exponent. $\displaystyle n(x^{n-1})$

$\displaystyle (x^2)^2 = x^4$

Then use power rule and you get

 $\displaystyle 4x^3$

Calculate the derivative of the function g(t).

$\displaystyle g(t)=5t^2-6cos(t)+sin(t)$

We need to recall a few rules in order to compute this derivative.

1) The derivative of a monomial can be found by multiplying the coefficent by the exponent, and decreasing the exponent by 1.

$\displaystyle 5t^2\rightarrow (2)5t=10t$

2) The derivative of cosine is negative sine.

$\displaystyle -6cos(t)\rightarrow(-)-6sin(t)=6sin(t)$

3) The derivative of sine is cosine.

$\displaystyle sin(t)\rightarrow cos(t)$

Put it all together to get our answer:

$\displaystyle g'(t)=10t+6sin(t)+cos(t)$

We cannot draw any conclusions about the derivative of $\displaystyle f(x)$ or its differentiablity; the expression we would want to talk about is$\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(h)}{h}$, not $\displaystyle \lim_{h \to 0} \frac{f(x+h)-f(h)}{x}$. Notice that the denominator of the first fraction is incorrect.