MHB Deriving F(x) from f(x*f(x^2)) with Given Conditions

[Umar]

F = f(x*f(x^2)), such that  f (4) = 6,  f '(4) = 1, and  f '(12) = 3. Find F '(2)

I know the format looks weird, but that's exactly how the function was written, which is why I'm not sure how to proceed with this one.

 

[Theia]

What ideas you have about how to do this problem? Or, do some rules of derivatives look promising?

 

[Umar]

What ideas you have about how to do this problem? Or, do some rules of derivatives look promising?

I was thinking of doing the product rule, but I just don't understand the xf(x^2) part of it. Now I'm also thinking the chain rule would be a better option actually. I'm just not sure how you would differentiate the x...

 

[Theia]

Yes, you'll need the product rule and the chain rule. ^^ The innerfunction is simply a function - and you need to use the chain rule again here.

For example if we have:

$$g = x(x - 1)^2$$,

we obtain

$$g' = (x - 1)^2 + x \cdot 2 (x-1)^1 \cdot 1 = \cdots$$.

Can you follow the procedure in your problem?

 

[Umar]

Yes, you'll need the product rule and the chain rule. ^^ The innerfunction is simply a function - and you need to use the chain rule again here.

For example if we have:

$$g = x(x - 1)^2$$,

we obtain

$$g' = (x - 1)^2 + x \cdot 2 (x-1)^1 \cdot 1 = \cdots$$.

Can you follow the procedure in your problem?

Yes I have tried but I just can't seem to work it out.

 

[MarkFL]

We are given:

$$F(x)=f\left(xf\left(x^2\right)\right)$$

Using the chain and product rules, we obtain:

$$F'(x)=f'\left(xf\left(x^2\right)\right)\left(xf'\left(x^2\right)(2x)+1\cdot f\left(x^2\right)\right)$$

Simplify:

$$F'(x)=f'\left(xf\left(x^2\right)\right)\left(2x^2f'\left(x^2\right)+f\left(x^2\right)\right)$$

What do you get when you let $x=2$?

 

[Umar]

It's alright, I ended up getting the answer, but thank you for outlining the process :)