How do you differentiate #f(x) = sin((− x^2 − 1)^2) *(x^2 − 9)^2# using the chain rule?

1 Answer

Answer:

Hey there!

To differentiate your function, you want to start by looking, overall, at what type of function you have. In short, you have a product (with chains inside). This could get a bit messy, haha!

Recall that chain rule looks like this:

If #f(x) = g(h(x)) # then,

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

Explanation:

First, start off with the product rule where if:

# f(x) = g(x)h(x)# then,

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

Note that the order of multiplication doesn't matter!

Starting off with your derivative:

Differentiate the first part: #sin((-x^2-1)^2)#:

The derivative of sin(x) is cos(x), but your "x" is much more complicated; it's a chain of functions and thus, by Chain Rule, you must multiply by the inside of the function. Therefore the derivative of the first part is:

# cos(-x^2-1)^2 * 2(-x^2-1)*(-2x) #

Note that you had to do chain rule on the # (-x^2-1)^2 # since that whole expression is being raised to a power. That's where the # -2x # came from. Also, to get the 2 in front, you do Power Rule since you have a function being raised to a power. Now just add in the second function "h(x)" unchanged as seen in the general product rule formula.

So far we have:

# f'(x) = (cos(-x^2-1)^2 * 2(-x^2-1)*(-2x))((x^2-9)^2) + ... #

Now we need to differentiate the second half of the equation. This is a simple chain rule problem.

To differentiate # (x^2-9)^2#:

By Chain Rule / Power Rule the derivative is:

# 2(x^2-9) * (2x)#

Now you just multiply by the first part of the function unchanged to get:

#(2(x^2-9)*2x)(sin((-x^2-1)^2)#

Finally, combine everything to get:

# f'(x) = (cos(-x^2-1)^2 * 2(-x^2-1)*(-2x))((x^2-9)^2) + (2(x^2-9)*2x)(sin((-x^2-1)^2) #

I know this was a bit messy, but hopefully everything was clear! Chain rule, combined with the other rules for derivatives can get quite complicated, it just requires practice! Hopefully this helps! :)