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Let \(y = \left(x+\sin\!\left(x\right)\right)^{4}.\)
Find \(g(x)\) and \(f(x)\) so that \(y = ( f \circ g ) ( x)\), and compute the derivative using the Chain Rule.

 

\(f(x) = \) 

\(g(x) = \) 

\((f \circ g) ' = \) 

 Mar 28, 2022
 #1
avatar+117857 
+1

g(x)=x+sinx

f(x)=x^4

 

There is a start

 Mar 28, 2022
 #2
avatar+73 
0

The Answers Are Below: 

 

\(f(x) = x^4\)

 

\(g(x) = x + sin(x)\)

 

\((f \circ g) ' = 4x^3+4(3x^2sin(x)+cos(x)x^3)+6(2xsin^2(x)+sin(2x)x^2)+4(sin^3(x)+3xsin^2(x)cos(x))+4sin^3(x)cos(x)\)

GAMEMASTERX40  Mar 30, 2022
 #3
avatar+117857 
+1

I really do not like the way this question is presented.  

Lets see if I can do better... 

\(g(x)=x+sinx\\ f(x)=x^4\\ y(x)=(f\circ g)(x)=[g(x)]^4=(x+sinx)^4 \)

 

now I have just referred to g(x) as g  etc but you have to remember that g, f and y are all functions of x.  

I have also stopped talking about f becasue I find it confusing.

This is probably not the most mathematically correct presentation but I think it is easier to understand.

 

\(y=(x+sinx)^4\\ let \;\; g=x+sinx\quad and\quad y=g^4\\ so\\ \frac{dg}{dx}=1+cosx \quad and\quad \frac{dy}{dg}=4g^3=4(x+sinx)^3\\ \frac{dy}{dx}=\frac{dy}{dg}*\frac{dg}{dx}\\ \qquad \text{notice that the dg's cancel each other out - There is a chain of cancelling}\\ \frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\ \frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\\)

 

Now look at the pattern of what has happened.  That is an easier way to use this.

 

 

 

LaTex:

y=(x+sinx)^4\\
let \;\; g=x+sinx\quad and\quad  y=g^4\\
so\\
\frac{dg}{dx}=1+cosx \quad and\quad  \frac{dy}{dg}=4g^3=4(x+sinx)^3\\
\frac{dy}{dx}=\frac{dy}{dg}*\frac{dg}{dx}\\
\qquad \text{notice that the dg's cancel each other out}\\
\frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\
\frac{dy}{dx}=4(x+sinx)^3*(1+cosx)\\

Melody  Mar 31, 2022

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