In Love With Differentiation

Calculus Level 5

If $f'(x) = g(x)$ and $g'(x) =f(x^2)$ , then $f''(x^3)$ can be expressed in the form

$\large k{x^a}f(x^b) + pxg(x^c)$

where $a$ , $b$ , $c$ , $k$ , and $p$ are natural numbers. Find the value of $k +p+a+b+c$ .

The answer is 28.

1 solution

Mukesh Ramanathan
May 28, 2019

f'( $x^{3}$ ) = $\frac{d}{dx}$ f( $x^{3}$ ) $\times$ 3 $x^{2}$ = g( $x^{3}$ ) $\times$ 3 $x^{2}$ using the chain rule

f''( $x^{3}$ ) = $\frac{d}{dx}$ (g( $x^{3}$ ) $\times$ 3 $x^{2}$ ) = 6 $x$ $\times$ g( $x^{3}$ )+3 $x^{2}$ $\times$ 3 $x^{2}$ f( $x^{6}$ ) using the product rule

=9 $x^{4}$ f( $x^{6}$ )+6 $x$ g( $x^{3}$ )

9+4+6+6+3= $\boxed{28}$

Does such a function exist other than $f(x) = 0$ ?

D G - 2 years ago

Hmmm....cant say

Gyandeep Kumar Yadav - 2 years ago

Solution is wrong. It will be f'(x^3)=g(x^3) by substituting x^3 instead of x in the question.

ALOK THAKUR - 2 years ago

No you will have to use the chain rule. For example let f(x)= x^2And g(x) = 2x and try using your substitution

Mukesh Ramanathan - 2 years ago

Wrong solution #Mukesh Ramanathan Correct Solution f'(x^3) = g(x^3), differentiating both sides wrt x => f"(x^3) × (3x^2) = (3x^2) × g'(x^3) => f"(x^3) = g'(x^3) = f(x^6)

Soumya Dasgupta - 2 years ago

The notation of this problem is confusing and misleading. It is unclear if you mean f(x^3)’’ or f’’(x^3). The two mean completely different things.

JD Money - 1 year, 12 months ago

In Love With Differentiation

The answer is 28.

1 solution

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