## Calculus Tutorial

*continuation of chain rule*

#### Intro

The chain rule in calculus is finding the derivative for two or more products of themselves. For example F(x)= (x+3)^4 where F(x)= G(H(x)) and G(x)= (x)^4, H(x)= x+3. Then you would use the chain rule to find the derivative. This rule is simple, yet it can be hard to differentiate the different individual functions. So for example the derivative F'(x)=G'(H(x))*H'(x). One easy way to remember is to think, derivative of the outside times derivative of the inside. Lets dive into it…..these are a lot harder than the intro tutorial.

#### Sample Problem

D(x)=(e^3x)^3

T(x)=(x-4x^2+13)^2/3

R(x)=17(x^5+3x^3+27)/15)^1/2

F(x)=4((2x^2+5x+10)/5)^5/6

find the derivatives

#### Solution

D(x)=(e^3x)^3

lets differentiate the functions

D'(x)=K'(Z(x))*Z'(x)

where K'(x)=3x^2 and Z'(x)= 3e^3x

so

D'(x)=3(Z(x))^2 * Z'(x)

D'(x)=3(e^3x)^2 * 3e^3x

T(x)=(x-4x^2+13)^2/3

lets differentiate the functions

T'(x)=U'(L(x))*L'(x)

where U'(x)=2/3(x)^(-1/3) and L'(x)= 1-8x

so

T'(x)=2/3(L(x))^(-1/3)*L'(x)

T'(x)=2/3(x-4x^2+13)^(-1/3)* (1-8x)

R(x)=17(x^5+3x^3+27)/15)^1/2

lets differentiate the functions

V(x)=17(x)^1/2 and W(x)=(x^5 +3x^3 +27)/15)

so for R'(x)

R'(x)=V'(W(x))*W'(x)

where V'(x)= 17/2(x)^(-1/2) and W'(x)= (5x^4 +9x^2)/15 =(x^4/3 + 3x^2/5)

so

R'(x)=17/2(W(x))^(-1/2) * W'(x)

R'(x)=17/2((x^5 +3x^3 +27)/15))^(-1/2) * (x^4/3 + 3x^2/5)

F(x)=4((2x^2+5x+10)/5)^5/6

lets differentiate the functions

G(x)=4(x)^5/6 and H(x)= (2x^2+5x+10)/5

so for F'(x)

F'(x)=G'(H(x))*H'(x)

where G'(x)= 4*(5/6)(x)^(-1/6) and H'(x)=(4x+5)/5= (4x/5 +1)

so

F'(x)= 20/6(H(x))^)(-1/6)*H'(x)

F'(x)=20/6((2x^2+5x+10)/5)^(-1/6)* (4x/5 +1)

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