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Question: Let \(f:\left( { - 1,1} \right) \to R\) be a differentiable function with \(f\left( 0 \right) = - 1\...

Let f:(1,1)Rf:\left( { - 1,1} \right) \to R be a differentiable function with f(0)=1f\left( 0 \right) = - 1 and f(0)=1f'\left( 0 \right) = 1. Let g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}. Then g(0)=g'\left( 0 \right) =
(A) 4 - 4
(B) 00
(C) 2 - 2
(D) 44

Explanation

Solution

A function g(x)g\left( x \right) in relation with another function f(x)f\left( x \right) is given in the question. At first, the derivative of the given function g(x)g\left( x \right) has to be derived. Then, by substituting the value of the function f(x)f\left( x \right) and derivative of the function f(x)f'\left( x \right) in the derived function g(x)g'\left( x \right), we can calculate the value of g(0)g'\left( 0 \right).

Formula used: ddx(x)=1 ddx(xn)=nxn1 ddx(n)=0  \dfrac{d}{{dx}}\left( x \right) = 1 \\\ \dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}} \\\ \dfrac{d}{{dx}}\left( n \right) = 0 \\\
Where,
xx is the variable
nn is the constant
ddx\dfrac{d}{{dx}} is the rate of change with respect to xx.

Complete step-by-step answer:
The function, g(x)=[f(2f(x)+2)]2g\left( x \right) = {\left[ {f\left( {2f\left( x \right) + 2} \right)} \right]^2}
The function of f(1,1)f\left( { - 1,1} \right), RR is the differentiable function.
f(0)=1f\left( 0 \right) = - 1, and
Derivative of f,f(0)=1f,\,f'\left( 0 \right) = 1

Complete step by step solution:
The function g(x)g\left( x \right) is considered as,
g\left( x \right) = {\left\\{ {f\left( {2f\left( x \right) + 2} \right)} \right\\}^2}
Differentiating g(x)g\left( x \right) with respect to xx,
g(x)=2[f(2f(x)+2)].f(2f(x)).2f(x)g'\left( x \right) = 2\left[ {f\left( {2f\left( x \right) + 2} \right)} \right].f'\left( {2f\left( x \right)} \right).2f'\left( x \right)
Simplifying the above function
g(x)=4[f(2f(x)+2)].f(2f(x)+2).f(x)g'\left( x \right) = 4\left[ {f\left( {2f\left( x \right) + 2} \right)} \right].f'\left( {2f\left( x \right) + 2} \right).f'\left( x \right)
Substituting the values of xx with 00 in order to find g(0)g'\left( 0 \right)
g(0)=4[f(2f(0)+2)].f(2f(0)+2).f(0)g'\left( 0 \right) = 4\left[ {f\left( {2f\left( 0 \right) + 2} \right)} \right].f'\left( {2f\left( 0 \right) + 2} \right).f'\left( 0 \right)
Substituting the value of f(0)f\left( 0 \right) as 1 - 1,
g(0)=4[f(2(1)+2)].f((2×1)+2).f(0) g(0)=4[f(2+2)].f(2+2).f(0)  g'\left( 0 \right) = 4\left[ {f\left( {2\left( { - 1} \right) + 2} \right)} \right].f'\left( {\left( {2 \times - 1} \right) + 2} \right).f'\left( 0 \right) \\\ g'\left( 0 \right) = 4\left[ {f\left( { - 2 + 2} \right)} \right].f'\left( { - 2 + 2} \right).f'\left( 0 \right) \\\
Further simplifying the above equation,
g(0)=4f(0).f(0).f(0)g'\left( 0 \right) = 4f\left( 0 \right).f'\left( 0 \right).f'\left( 0 \right)
Substituting the value of f(0)f'\left( 0 \right) as 11,
g(0)=4×1×1×1 g(0)=4  g'\left( 0 \right) = 4 \times - 1 \times 1 \times 1 \\\ g'\left( 0 \right) = - 4 \\\
Hence, the value of g(0)g'\left( 0 \right) is obtained as 4 - 4.

Thus, the option (A) is correct.

Note: Differentiation and derivatives approaches are first provided by Isaac Newton and Gottfried Wilhelm Leibniz. This is called the modern development of calculus. Derivatives are used in the process of modelling of moving objects. It is also used for calculating profit or loss in the business, to find speed or distance covered, to find temperature difference etc. Differentiation deals with finding its derivative of one variable with respect to the other. There are three derivatives which are basics in nature. It includes algebraic functions, exponential functions and trigonometric functions. For example, if xx and yy are the two variables, then the rate of change of xx with respect to yy is dxdy\dfrac{{dx}}{{dy}}.