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

Hint: 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.