Question

If $g$ is twice differentiable function and $f\left( x \right) = xg\left( {{x^2}} \right)$, how do you find in terms of $g$, $g'$, and $g''$?

Answer 1

Here in this question, we have to differentiate the given function $f\left( x \right) = xg\left( {{x^2}} \right)$ and express of in terms of $g$, $g'$, and $g''$. First, we need to use the product rule for differentiating the function $f\left( x \right)$ and lateral by chain rule also differentiate the function $g\left( x \right)$ and second time differentiate the function $f\left( x \right)$ and by further simplification using the standard differentiating formula to get the required solution.

Complete step by step answer:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
$\Rightarrow \,\,\,\,f\left( x \right) = xg\left( {{x^2}} \right)$---------- (1)
Differentiate function $f\left( x \right)$ with respect to x
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {xg\left( {{x^2}} \right)} \right)$
Now, we need to use the product rule of differentiation $\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{d}{{dx}}\left( v \right) + v\dfrac{d}{{dx}}\left( u \right)$, then
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = x \cdot \dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) + g\left( {{x^2}} \right) \cdot \dfrac{d}{{dx}}\left( x \right)$
As we know, the standard formula: $\dfrac{{dx}}{{dx}} = 1$, and represent $\dfrac{d}{{dx}}\left( {f\left( x \right)} \right) = f'\left( x \right)$.
$\Rightarrow \,\,\,\,f'\left( x \right) = x \cdot \dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) + g\left( {{x^2}} \right) \cdot \left( 1 \right)$
$\Rightarrow \,\,\,\,f'\left( x \right) = x \cdot \dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) + g\left( {{x^2}} \right)$
Now, then by the chain rule, we have:
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) = g'\left( {{x^2}} \right)\dfrac{d}{{dx}}\left( {{x^2}} \right)$
As we know the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, then
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) = g'\left( {{x^2}} \right)2x$
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right) = 2xg'\left( {{x^2}} \right)$----------(2)
Substitute (2) in equation (1), then
$\Rightarrow \,\,\,\,f'\left( x \right) = x \cdot \left( {2xg'\left( {{x^2}} \right)} \right) + g\left( {{x^2}} \right)$
$\Rightarrow \,\,\,\,f'\left( x \right) = 2{x^2}g'\left( {{x^2}} \right) + g\left( {{x^2}} \right)$--------(3)
Again, differentiating equation (3) a second time using product rule, then
$\Rightarrow \,\,\,\,\dfrac{d}{{dx}}\left( {f'\left( x \right)} \right) = \dfrac{d}{{dx}}\left( {2{x^2}g'\left( {{x^2}} \right) + g\left( {{x^2}} \right)} \right)$
Represent $\dfrac{d}{{dx}}\left( {f'\left( x \right)} \right) = f''(x)$.
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) + g'\left( {{x^2}} \right)\dfrac{d}{{dx}}\left( {2{x^2}} \right) + \dfrac{d}{{dx}}\left( {g\left( {{x^2}} \right)} \right)$
using the formula $\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}$, then
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) + g'\left( {{x^2}} \right)2\left( {2x} \right) + g'\left( {{x^2}} \right)\left( {2x} \right)$
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) + 4x\,g'\left( {{x^2}} \right) + 2x\,g'\left( {{x^2}} \right)$
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) + 6x\,g'\left( {{x^2}} \right)$--------(4)
And, again by the chain rule, we have:
$\Rightarrow \,\,\,\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) = g''\left( {{x^2}} \right)\,\dfrac{d}{{dx}}\left( {{x^2}} \right)$
$\Rightarrow \,\,\,\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) = g''\left( {{x^2}} \right)\,\,2x$
$\Rightarrow \,\,\,\dfrac{d}{{dx}}\left( {g'\left( {{x^2}} \right)} \right) = 2x\,g''\left( {{x^2}} \right)$-----------(5)
Substitute equation (5) in (4), then
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\left( {2x\,g''\left( {{x^2}} \right)} \right) + 6x\,g'\left( {{x^2}} \right)$
$\Rightarrow \,\,\,\,f''\left( x \right) = 2{x^2}\left( {2x\,g''\left( {{x^2}} \right)} \right) + 6x\,g'\left( {{x^2}} \right)$
On simplification, we get
$\Rightarrow \,\,\,\,f''\left( x \right) = 4{x^3}g''\left( {{x^2}} \right) + 6x\,g'\left( {{x^2}} \right)$
Hence, it’s a required solution.

Note: When solving these types of questions, the student must know about the standard differentiation formulas. If the function is a product of two terms and the both terms are the function of x the we use the product rule of differentiation to the function and behaviour of the chain rule while solving.