How to differentiate y=cos4(x2 + ex)?

[futurept]

Homework Statement

The problem is to differentiate the function.
y=cos⁴(x² + e^x)

Homework Equations

cos(x)'= -sin(x), (xⁿ)'=n*x^n-1, (e^x)'=e^x*x'

The Attempt at a Solution

Thought it would be the chain rule. Here's what I came up with:

y'=4cos³(-sin(x² + e^x)(2x + e^x)

Is this right? If not, any suggestions on what I did wrong?

 

[gabbagabbahey]

y'=4cos³(-sin(x² + e^x)(2x + e^x)

Is this right? If not, any suggestions on what I did wrong?

Is this a typo? What are you taking the cosine of?

 

[futurept]

no it's not a typo. my answer was y'= 4cos³(-sin(x² + e^x))*(2x + e^x)

 

[futurept]

I did forget to say that the problem says not to simplify.

 

[chislam]

You're close but take another look at the chain rule.

f(x) = h(g(x))

f'(x) = h'(g(x)) * g'(x)

Your answer follows the following incorrect differentiation.

f'(x) = h'(g'(x))

But you only messed up for one part of the chain. The innermost function was differentiated correctly.

 

[gabbagabbahey]

no it's not a typo. my answer was y'= 4cos³(-sin(x² + e^x))*(2x + e^x)

In that case it is incorrect.

Try using the substitution $u=x^2+e^x$ along with the chain rule...What is $$\frac{d}{dx}\cos^4(u)$$?

 

[futurept]

so h(x)=cos⁴(x²+e^x)

and g(x)= x² + e^x

but would the cos⁴ part be another function by itself? So then it would be f(g(h(x))).

and by definition of the chain rule it would be f '(g(h(x)))*g'(h(x))*h'(x)?

 

[gabbagabbahey]

so h(x)=cos⁴(x²+e^x)

and g(x)= x² + e^x

but would the cos⁴ part be another function by itself? So then it would be f(g(h(x))).

and by definition of the chain rule it would be f '(g(h(x)))*g'(h(x))*h'(x)?

hmmm this doesn't make much sense.

Instead, let's call $g(x) \equiv \cos^4(x)$ and $h(x)\equiv x^2+e^x$ then $f(x)\equiv\cos^4(x^2+e^x)=g(h(x))$

...follow?

Now, what does the chain rule give you for $f'(x)$?

 

[futurept]

yeah, i think i follow. so from your definition, f'(x)=g'(h(x))*h'(x)?

 

[futurept]

so, (cos(x^2+e^x))^4

f'(x)= 4cos(x^2+e^x)^3*-sin(x^2+e^x)*(2x+e^x)

 

[gabbagabbahey]

so, (cos(x^2+e^x))^4

f'(x)= 4cos(x^2+e^x)^3*-sin(x^2+e^x)*(2x+e^x)

Yes, much better!

 

[futurept]

you guys are awesome