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# Let y = (x + sin(x))^4. Find g(x) and f(x) so that y = (f o g)(x), and compute the derivative using the Chain Rule.

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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

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g(x)=x+sinx

f(x)=x^4

There is a start

Mar 28, 2022
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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
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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\\