Questions & Answers

Question

Answers

A. \[\dfrac{{ - 1}}{3}\]

B. \[\dfrac{1}{{13}}\]

C. \[\dfrac{1}{3}\]

D. \[\dfrac{{ - 1}}{{13}}\]

Answer
Verified

* Chain rule of differentiation is given as \[\dfrac{d}{{dx}}f(g(x)) = f'(g(x)) \times g'(x)\] where \[f'(g(x))\] is differentiation of f and \[g'(x)\] is differentiation of g with respect to x.

We have the function \[f(x) = {x^2} - x + 5\]

We find the differentiation of \[f(x)\]

\[ \Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}({x^2} - x + 5)\]

\[ \Rightarrow \dfrac{d}{{dx}}f(x) = (2x - 1)\] … (1)

Now we know that \[g(x)\] is an inverse function of \[f(x)\].

Then we can write that when g is applied to f we get the value x

\[ \Rightarrow g(f(x)) = x\]

Now we differentiate both sides of the equation with respect to x

\[ \Rightarrow \dfrac{d}{{dx}}g(f(x)) = \dfrac{d}{{dx}}(x)\]

Using chain rule on LHS of the equation and differentiating RHS of the equation.

\[ \Rightarrow g'(f(x)).f'(x) = 1\]

Dividing both sides of the equation by \[f'(x)\]

\[ \Rightarrow \dfrac{{g'(f(x)).f'(x)}}{{f'(x)}} = \dfrac{1}{{f'(x)}}\]

Cancel the same terms from numerator and denominator.

\[ \Rightarrow g'(f(x)) = \dfrac{1}{{f'(x)}}\]

Substituting the value from equation (1), we get

\[ \Rightarrow g'(f(x)) = \dfrac{1}{{(2x - 1)}}\] … (2)

We know that \[g(f(x)) = x\], We have to find the value of \[g'(7)\]. We find the value of \[g(7)\], i.e.

\[ \Rightarrow f(x) = 7\]

Substitute the value of \[f(x) = {x^2} - x + 5\]

\[ \Rightarrow {x^2} - x + 5 = 7\]

Shift all values to one side of the equation

\[

\Rightarrow {x^2} - x + 5 - 7 = 0 \\

\Rightarrow {x^2} - x - 2 = 0 \\

\]

Now we factorize the equation by writing \[ - x = - 2x + x\]

\[ \Rightarrow {x^2} - 2x + x - 2 = 0\]

Take x common from first two terms and 1 common from last two terms

\[ \Rightarrow x(x - 2) + 1(x - 2) = 0\]

Pair up the factors

\[ \Rightarrow (x - 2)(x + 1) = 0\]

Now we can equate both the factors equal to zero.

Firstly \[x - 2 = 0\]

Shift the constant to one side of the equation.

\[ \Rightarrow x = 2\]

Secondly, \[x + 1 = 0\]

Shift the constant to one side of the equation.

\[ \Rightarrow x = - 1\]

So, \[x = 2,x = - 1\]

But we have a condition \[x > \dfrac{1}{2}\] i.e. \[x > 0.5\]

So we have to reject the value of \[x = - 1\].

So, our value of \[x = 2\].

Now we substitute the value in equation (2)

\[

\Rightarrow g'(f(2)) = \dfrac{1}{{(2 \times 2 - 1)}} \\

\Rightarrow g'(f(2)) = \dfrac{1}{{(4 - 1)}} \\

\Rightarrow g'(f(2)) = \dfrac{1}{3} \\

\]

So, the value of \[g'(7) = \dfrac{1}{3}\]

Students many times make mistake of writing inverse of function by taking the reciprocal of the function, i.e. if \[f(x) = 2x + 5\], then they write \[{f^{ - 1}}(x) = \dfrac{1}{{2x + 5}}\] which is wrong.

×

Sorry!, This page is not available for now to bookmark.