Question

If we have the function as $f\left( x \right) = \sin x$, $g\left( x \right) = {x^2}$ and $h\left( x \right) = {\log _e}x$. If $F\left( x \right) = hogof\left( x \right)$, then $F''\left( x \right)$ is equal to:
(A). $a\cos e{c^3}x$
(B). $2{\cot ^2}x - 4{x^2}\cos e{c^2}{x^2}$
(C). $2{\cot ^2}x$
(D). $- 2\cos e{c^2}x$

Answer 1

In the given question, we are required to find the second derivative of a composite function. We are given the individual functions in the problem itself and we have to first calculate a composite function using the original functions and then find out the second derivative using the chain rule of differentiation. This question requires us to have the knowledge of basic and simple algebraic rules and operations such as substitution, addition, multiplication, subtraction and many more like these. A thorough understanding of functions and differentiation can be of great significance.

Complete step-by-step solution:
In the given question, we are given the functions $f\left( x \right) = \sin x$, $g\left( x \right) = {x^2}$ and $h\left( x \right) = {\log _e}x$.
So, we have to first find the composite function $F\left( x \right) = hogof\left( x \right)$
The composite function to be calculated is F\left( x \right) = h\left[ {g\left\\{ {f\left( x \right)} \right\\}} \right].
So, we have to change the value of the variable in the original function in order to find the composite function.
Now, we know that $f\left( x \right) = \sin x$. So, we substitute the value of $f\left( x \right)$ as $\sin x$. Hence, we get,
\Rightarrow F\left( x \right) = h\left[ {g\left\\{ {\sin x} \right\\}} \right]
Now, to carry on the method of evaluating the composite function, we have to change the variable in the function $g\left( x \right) = {x^2}$ from x to $\sin x$.
So, we get,
$\Rightarrow F\left( x \right) = h\left[ {{{\sin }^2}x} \right]$
Again, substituting the value of variable in function $h\left( x \right) = {\log _e}x$ as ${\sin ^2}x$. So, we get,
$\Rightarrow F\left( x \right) = {\log _e}\left( {{{\sin }^2}x} \right)$
So, the composite function $F\left( x \right) = hogof\left( x \right)$ is equivalent to $F\left( x \right) = {\log _e}\left( {{{\sin }^2}x} \right)$.
Now, we have to find the second derivative of the resultant function.
Differentiating the function once with respect to x, we get,
$\dfrac{d}{{dx}}F\left( x \right) = \dfrac{d}{{dx}}\left[ {{{\log }_e}\left( {{{\sin }^2}x} \right)} \right]$
Using the chain rule of differentiation $\dfrac{d}{{dx}}\left( {f\left( {g\left( x \right)} \right)} \right) = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)$, we get,
$\Rightarrow F'\left( x \right) = \dfrac{1}{{\left( {{{\sin }^2}x} \right)}} \times \dfrac{{d\left( {{{\sin }^2}x} \right)}}{{dx}}$
Since we know the derivative of $\log x$ and $\sin x$ as $\dfrac{1}{x}$ and $\cos x$. Also, power rule of differentiation is $\dfrac{{d\left( {{x^n}} \right)}}{{dx}} = n{x^{n - 1}}$.
$\Rightarrow F'\left( x \right) = \dfrac{1}{{\left( {{{\sin }^2}x} \right)}} \times 2\sin x \times \dfrac{{d\left( {\sin x} \right)}}{{dx}}$
$\Rightarrow F'\left( x \right) = \dfrac{1}{{\left( {{{\sin }^2}x} \right)}} \times 2\sin x \times \cos x$
Simplifying the expression, we get,
$\Rightarrow F'\left( x \right) = 2\dfrac{{\cos x}}{{\sin x}}$
We know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$. So, we get,
$\Rightarrow F'\left( x \right) = 2\cot x$
Now, differentiating the function once more with respect to x,
$\Rightarrow F''\left( x \right) = \dfrac{d}{{dx}}\left( {2\cot x} \right)$
Now, we know that the derivative of $\cot x$ with respect to x is $- \cos e{c^2}x$.
$\Rightarrow F''\left( x \right) = - 2\cos e{c^2}x$
So, the value of $F''\left( x \right)$ is $- 2\cos e{c^2}x$.
Hence, option (D) is correct.

Note: Such questions that require just a simple change of variable can be solved easily by keeping in mind the algebraic rules such as substitution and transposition. The chain rule of differentiation involves differentiating a composite by introducing new unknowns to ease the process and examine the behaviour of function layer by layer.