MHB Find xf^7(x)''(0): Chain Rule Explanation

[needOfHelpCMath]

Assume we known that f(0) = 1 and f'(0)=2
Find $$xf^7(x)''(0)$$

Will chain rule work here?

is the $$ u=xf^7$$

and $$ y = u^7$$

I don't know if I am going in the right direction.

 

[Opalg]

Assume we known that f(0) = 1 and f'(0)=2
Find $$xf^7(x)''(0)$$

Will chain rule work here?

is the $$ u=xf^7$$

and $$ y = u^7$$

I don't know if I am going in the right direction.

As you have noticed, this is an ambiguously worded question. I think that what it is asking for is $$\frac {d^2}{dx^2}\bigl(x(f(x))^7\bigr)$$ (evaluated at $x=0$).

 

[needOfHelpCMath]

As you have noticed, this is an ambiguously worded question. I think that what it is asking for is $$\frac {d^2}{dx^2}\bigl(x(f(x))^7\bigr)$$ (evaluated at $x=0$).

this is what I got from solving the problem:

For the first derivative

$$ d/dx[f^7x^8]$$
$$ f^7*d/dx[x^8]$$
$$8f^7x^7$$Second derivative:

$$d/dx8f^7x^7$$
$$8f^7*d/dx[x^7]$$
$$=8*7x^6f^7$$
$$56f^7x^6$$

 

[MarkFL]

I would let:

$$g(x)=xf^7(x)$$

Differentiate w.r.t $x$, using the product and chain rules:

$$g'(x)=x(7f^6(x)f'(x))+(1)f^7(x)=7xf^6(x)f'(x)+f^7(x)=f^6(x)(7xf'(x)+f(x))$$

Now, differentiate again:

$$g''(x)=f^6(x)(7xf''(x)+7f'(x)+f'(x))+6f^5(x)f'(x)(7xf'(x)+f(x))=f^5(x)(f(x)(7xf''(x)+8f'(x))+6f'(x)(7xf'(x)+f(x)))$$

$$g''(x)=7f^5(x)(xf(x)f''(x)+2f(x)f'(x)+6xf'^2(x))$$

Now, can you find $g''(0)$?