Step 1: Define $\displaystyle f(x),g(x)$.

$\displaystyle f(x)=x+2, g(x)=x$

Step 2: Find $\displaystyle f'(x),g'(x)$.

$\displaystyle f'(x)=1,g'(x)=1$

Step 3: Plug in the functions/values into the formula for quotient rule: $\displaystyle \frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{[g(x)]^2}$

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

The derivative of the expression is $\displaystyle -\frac {2}{x^2}$

This question yields to application of the quotient rule:

 $\displaystyle \left(\frac{f}{g}\right)'=\frac{f'g-fg'}{g^2}$

So find $\displaystyle f'$ and $\displaystyle g'$ to start:

$\displaystyle f(x)=5x^3$

$\displaystyle f'(x)=15x^2$

$\displaystyle g(x)=e^x$

$\displaystyle g'(x)=e^x$

$\displaystyle \left(\frac{f}{g}\right)'=\frac{15x^2\cdot e^x-5x^3\cdot e^x}{(e^x)^2}$

$\displaystyle \left(\frac{f}{g}\right)'=\frac{e^x(15x^2-5x^3)}{(e^x)(e^x)}=\frac{15x^2-5x^3}{e^x}$

So our answer is:

$\displaystyle \frac{15x^2-5x^3}{e^x}$

Finding the First Derivative:

Step 1: Define $\displaystyle f(x),g(x)$

$\displaystyle f(x)=x^3-2x, g(x)=x^5+4x^2$

Step 2: Find $\displaystyle f'(x), g'(x)$

$\displaystyle f'(x)=3x^2-2, g'(x)=5x^4+8x$

Step 3: Plug in all equations into the quotient rule formula: $\displaystyle \frac {f'g-g'f}{g^2}$

$\displaystyle \frac {(3x^2-2)(x^5+4x^2)-[(5x^4+8x)(x^3-2x)]}{[(x^5+4x^2)]^2}$

Step 4: Simplify the fraction in step 3:

$\displaystyle \frac {3x^7+12x^4-2x^5-8x^2-[5x^7-10x^5+8x^4-16x^2]}{x^{10}+8x^7+16x^4}$

$\displaystyle =\frac {3x^7+12x^4-2x^5-8x^2-5x^7+10x^5-8x^4+16x^2}{x^{10}+8x^7+16x^4}$
$\displaystyle =\frac {-2x^7+8x^5+4x^4+8x^2}{x^{10}+8x^7+16x^4}$

Step 5: Factor an $\displaystyle x^2$ out from the numerator and denominator. Simplify the fraction..

$\displaystyle \frac {x^2(-2x^5+8x^3+4x^2+8)}{x^2(x^8+8x^5+16x^2)}=\frac {-2x^5+8x^3+4x^2+8}{x^8+8x^5+16x^2}$
We have found the first derivative..

Finding Second Derivative:

Step 6: Find $\displaystyle f(x),g(x)$ from the first derivative function

$\displaystyle f(x)=-2x^5+8x^3+4x^2+8, g(x)=x^8+8x^6+16x^2$

Step 7: Find $\displaystyle f'(x), g'(x)$

$\displaystyle f'(x)=-10x^4+24x^2+8x, g'(x)=8x^7+40x^4+32x$

Step 8: Plug in the expressions into the quotient rule formula: $\displaystyle \frac {f'g-g'f}{g^2}$

$\displaystyle \frac {(-10x^4+24x^2+8x)(x^8+8x^5+16x^2)-[(8x^7+40x^4+32x)(-2x^5+8x^3+4x^2+8)]}{[(x^8+8x^6+16x^2)]^2}$

Step 9: Simplify:

$\displaystyle \frac {-10x^{12}-80x^9-160x^6+24x^{10}+192x^7+384x^4-...-64x^6-64x^{10}-...-256x}{[(x^8+8x^6+16x^2)]^2}$

I put "..." because the numerator is very long. I don't want to write all the terms...

Step 10: Combine like terms:

$\displaystyle \frac {-26x^{12}-88x^{10}-130x^9-296x^7-308x^6-680x^4-256x}{x^{16}+16x^{13}+96x^{10}+256x^7+256x^4}$

Step 11: Factor out $\displaystyle x$ and simplify:

Final Answer: $\displaystyle \frac {-26x^{11}-88x^9-130x^8-296x^6-308x^5-680x^3-256}{x^{15}+16x^{12}+96x^9+256x^6+256x^3}$.

This is the second derivative.


The answer is None of the Above. The second derivative is not in the answers...

This question requires application of multiple chain rules.  There are 2 inner functions in $\displaystyle y=cos(sin(2x))$, which are $\displaystyle sin(2x)$ and $\displaystyle 2x$.  

The brackets are to identify the functions within the function where the chain rule must be applied.

Solve the derivative.

$\displaystyle \frac{dy}{dx}= -sin([sin(2x)]) \cdot (cos[2x]))\cdot 2=-2cos(2x)sin(sin(2x))$

The sine of sine of an angle cannot be combined to be sine squared.

Therefore, the answer is: 

$\displaystyle -2cos(2x)sin(sin(2x))$

Recall chain rule for this problem

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

So if we are given the following,

$\displaystyle h(x)=(3x^3-4x)^{33}$

We can think of it like this

$\displaystyle f(x)=3x^3-4x$

$\displaystyle f'(x)=9x^2-4$

$\displaystyle g(x)=x^{33}$

$\displaystyle g'(x)=33x^{32}$

$\displaystyle (f(g(x)))'=(9x^2-4)(33)(3x^2-4x)^{32}$

Clean it up a bit to get:

$\displaystyle (f(g(x)))'=(297x^2-132)(3x^2-4x)^{32}$

Chain Rule:$\displaystyle Given: f(x)*g(y)$

$\displaystyle \frac{\mathrm{dy} }{\mathrm{d} x}=f(x)g'(y)+g(y)f'(x)$

For this problem

$\displaystyle f(x)=3x^2$

$\displaystyle f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}3x^2=3(2)x^{2-1}=6x$

 $\displaystyle g(y)=8y^3$

$\displaystyle g'(y)=\frac{\mathrm{d} }{\mathrm{d} x}8y^3=8(3)y^{3-1}=24y^2$

Plug the values into the Chain Rule formula and simplify:

$\displaystyle \frac{\mathrm{dy} }{\mathrm{d} x}(3x^2*8y^3)=3x^2*24y^2+8y^3*6x$

Step 1: Try plugging in $\displaystyle x=2$ into the denominator of the function. We want to make sure that the bottom does not become $\displaystyle 0$...

$\displaystyle 2^3-8=8-8=0$.. We got zero, and we cannot have zero in the denominator. So, we must try and factor the function (numerator and denominator):

Step 2: Factor:

$\displaystyle \frac {(x+2)(x-2)}{(x-2)(x^2+2x+4)}$

Step 3: Reduce:

$\displaystyle \frac {x+2}{x^2+2x+4}$

Step 4: Now that we got rid of the factor that made the denominator zero, we know that this function has a limit.

Step 5: Plug in $\displaystyle x=2$ into the reduced factor form:

Simplify as much as possible...

$\displaystyle \frac {2+2}{2^2+2(2)+4}=\frac {4}{12}=\frac {1}{3}$

The limit of this function as x approaches $\displaystyle 2$ is $\displaystyle \frac {1}{3}$

The first step is to differentiate both sides with respect to $\displaystyle x$:

$\displaystyle 3y^2 \ \frac{\mathrm{d} }{\mathrm{d} x}+12x \ \frac{\mathrm{d} }{\mathrm{d} x}-6\cos x \ \frac{\mathrm{d} }{\mathrm{d} x}=-4y^3\frac{\mathrm{d} }{\mathrm{d} x}$

Note: Those that are functions of $\displaystyle y$ can be differentiated with respect to $\displaystyle y$, just remember to mulitply it by $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$

$\displaystyle 6y \ \frac{\mathrm{d}y }{\mathrm{d} x}+12+6\sin x =-12y^2\frac{\mathrm{d} y}{\mathrm{d} x}$

Now we can solve for $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$:

$\displaystyle 6y \ \frac{\mathrm{d}y }{\mathrm{d} x}+12 y^2\frac{\mathrm{d} y}{\mathrm{d} x}=-6\sin x -12$

$\displaystyle (6y +12 y^2) \ \frac{\mathrm{d} y}{\mathrm{d} x}=-6(\sin x +2)$

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-6(\sin x +2)}{6(y +2 y^2)}$

$\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}= \frac{-(\sin x +2)}{y +2 y^2}$

Our first step would be to differentiate both sides with respect to $\displaystyle x$:

$\displaystyle x^2 \frac{\mathrm{d} }{\mathrm{d} x}+2y^3 \frac{\mathrm{d} }{\mathrm{d} x}=8 \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of $\displaystyle y$ can be differentiated with respect to $\displaystyle y$, just remember to multiply by  $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$.

$\displaystyle 2x + 6y^2 \frac{\mathrm{d}y }{\mathrm{d} x}=0$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-2x}{6y^2}$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}=\frac{-x}{3y^2}$

Our first step is to differentiate both sides with respect to $\displaystyle x$:

$\displaystyle \cos y \ \frac{\mathrm{d} }{\mathrm{d} x}+ 3x^3 \frac{\mathrm{d} }{\mathrm{d} x}+4y \frac{\mathrm{d} }{\mathrm{d} x}=\sin x \frac{\mathrm{d} }{\mathrm{d} x}$

The functions of $\displaystyle y$ can by differentiated with respect to $\displaystyle y$, just remember to multiply them by $\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}$: 

$\displaystyle -\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+ 9x^2 +4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x$

$\displaystyle -\sin y \ \frac{\mathrm{d}y }{\mathrm{d} x}+4 \frac{\mathrm{d}y }{\mathrm{d} x}=\cos x-9x^2$

$\displaystyle (-\sin y +4 )\frac{\mathrm{d}y }{\mathrm{d} x}=(\cos x-9x^2)$

$\displaystyle \frac{\mathrm{d}y }{\mathrm{d} x}= \frac{(\cos x-9x^2)}{(4-\sin y )}$