Question

If $\frac{{d(f(x))}}{{dx}} = \frac{1}{{1 + {x^2}}}$, then the \[\frac{{d(f({x^3}))}}{{dx}}\]is

Answer 1

Hint: The derivative of function f(x) is given. We can find the function f(x) by integrating both sides as follows:
$ \Rightarrow \int {\frac{{d(f(x))}}{{dx}}dx} = \int {\frac{1}{{1 + {x^2}}}dx} $, after integration the value of f(x) will be obtained the value of f(x).
To find the value of \[\frac{{d(f({x^3}))}}{{dx}}\], replace the x with ${x^3}$ and then take derivative of that function.

Complete step-by-step answer:
We are given with the derivation of a function i.e.
$ \Rightarrow $$\frac{{d(f(x))}}{{dx}} = \frac{1}{{1 + {x^2}}}$ …………… (1)
We have to find the value of \[\frac{{d(f({x^3}))}}{{dx}}\]as we don’t know the given function f(x).
First of all, we have to find the function f(x). Function f(x) can be obtained by integrating the equation (1) on both sides we get,
We will again apply formula on adjoint A i.e. adjoint of adjoint A is given by
$ \Rightarrow \int {\frac{{d(f(x))}}{{dx}}dx} = \int {\frac{1}{{1 + {x^2}}}dx} $
As $\frac{{d{{\tan }^{ - 1}}x}}{{dx}} = \frac{1}{{1 + {x^2}}}$and integration is opposite of differentiation, so
 $ \Rightarrow \int {\frac{1}{{1 + {x^2}}}dx} = {\tan ^{ - 1}}x + c$
$
   \Rightarrow f(x) = \int {\frac{{d(f(x))}}{{dx}}dx} = \int {\frac{1}{{1 + {x^2}}}dx} \\
   \Rightarrow f(x) = {\tan ^{ - 1}}x + c \\
    \\
$ ……………………. (2)
Replace the x term in equation (2) with ${x^3}$, we get
 $
   \Rightarrow f(x) = {\tan ^{ - 1}}x + c \\
   \Rightarrow f({x^3}) = {\tan ^{ - 1}}{x^3} + c \\
    \\
$, where c is a constant ……………………. (3)
At last we will differentiate the equation (3) on both sides, we get
$
   \Rightarrow \frac{{{d^{}}f({x^3})}}{{dx}} = \frac{{d({{\tan }^{ - 1}}{x^3} + c)}}{{dx}} \\
   \Rightarrow \frac{{{d^{}}f({x^3})}}{{dx}} = \frac{{d{{\tan }^{ - 1}}{x^3}}}{{dx}} + \frac{{d(c)}}{{dx}} \\
   \Rightarrow \frac{{{d^{}}f({x^3})}}{{dx}} = \frac{1}{{1 + {{({x^3})}^2}}}(\frac{{d{x^3}}}{{dx}}) \\
   \Rightarrow \frac{{{d^{}}f({x^3})}}{{dx}} = \frac{1}{{1 + {{({x^3})}^2}}}(3{x^2}) \\
   \Rightarrow \frac{{{d^{}}f({x^3})}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}} \\
$
The value of $\frac{{{d^{}}f({x^3})}}{{dx}} = \frac{{3{x^2}}}{{1 + {x^6}}}$
  So, the required answer is $\frac{{3{x^2}}}{{1 + {x^6}}}$.

Note: While doing integration we also add constant in the answer why? Because as we do differentiation of a constant then it is zero. Hence while doing integration we also consider that value by adding a constant to it.
Another common mistake done by students can be from $\frac{{{d^{}}f({x^3})}}{{dx}} = \frac{{d{{\tan }^{ - 1}}{x^3}}}{{dx}} + \frac{{d(c)}}{{dx}}$. In this while integrating the tangent function, the differentiate of ${x^3}$multiplied by differentiation of tangent as it is also a function in tangent function. Generally, students forget to do that. i.e.
$ \Rightarrow \frac{{d(f(f(x)))}}{{dx}} = f'(f(x)) \times f'(x)$