Question

If $8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5$and $y = {x^2}f(x)$, then what is $\dfrac{{dy}}{{dx}}$at $x = - 1$?
A. $0$
B. $\dfrac{1}{{14}}$
C. $- \dfrac{1}{{14}}$
D. $1$

Answer 1

Firstly, we will find the value of $f(x)$.For that we will first, replace $x$ by $\dfrac{1}{x}$ in the given equation $8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5$ and then we will obtain two equations in terms of $f(x)$ and $f\left( {\dfrac{1}{x}} \right)$. We will then solve the two equations for $f(x)$. After solving for $f(x)$, we will put that value of $f(x)$ in the given condition $y = {x^2}f(x)$ to obtain the value of $y$. Now, we will get the value of $y$ in terms of $x$i.e. we will get $y$ as a function of $x$. After obtaining the value of $y$ in terms of $x$, we will differentiate the value of $y$we will obtain in terms of $x$with respect to $x$. Now, as we have to find the value at $x = - 1$, we will substitute $x = - 1$in the value of $\dfrac{{dy}}{{dx}}$ obtained.

Complete step by step answer:
We are given,
$8f(x) + 6f\left( {\dfrac{1}{x}} \right) = x + 5 - - - - - - (1)$
Now, Replacing $x$ by $\dfrac{1}{x}$ in (1), we get
$\Rightarrow 8f\left( {\dfrac{1}{x}} \right) + 6f(x) = \dfrac{1}{x} + 5 - - - - - (2)$
Multiplying first equation by 8,
(1) $\times 8 \Rightarrow 8 \times \left[ {\left( {8f(x) + 6f\left( {\dfrac{1}{x}} \right)} \right) = x + 5} \right]$
$= 64f(x) + 48f\left( {\dfrac{1}{x}} \right) = 8x + 40 - - - - - (3)$

Multiplying second equation by 6,
(2) $\times 6 \Rightarrow 6 \times \left[ {8f\left( {\dfrac{1}{x}} \right) + 6f(x) = \dfrac{1}{x} + 5} \right]$
$48f\left( {\dfrac{1}{x}} \right) + 36f(x) = \dfrac{6}{x} + 30 - - - - - (4)$
Now, subtracting (4) from (3), we get
$\Rightarrow 64f(x) + 48f\left( {\dfrac{1}{x}} \right) - \left[ {48f\left( {\dfrac{1}{x}} \right) + 36f(x)} \right] = 8x + 40 - \left( {\dfrac{6}{x} + 30} \right)$
Opening the brackets, we get
$\Rightarrow 64f(x) + 48f\left( {\dfrac{1}{x}} \right) - 48f\left( {\dfrac{1}{x}} \right) - 36f(x) = 8x + 40 - \dfrac{6}{x} - 30$

Cancelling like terms with opposite signs, we get,
$\Rightarrow 28f(x) = 8x - \dfrac{6}{x} + 10$
So, we get the function as,
$f(x) = \dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)$
Now, we have, $f(x) = \dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)$
Substituting the value of $f(x)$in $y = {x^2}f(x)$, we get
$\Rightarrow y = {x^2}\left[ {\dfrac{1}{{28}}\left( {8x - \dfrac{6}{x} + 10} \right)} \right]$
$\Rightarrow y = \dfrac{{{x^2}}}{{28}}\left[ {8x - \dfrac{6}{x} + 10} \right]$
Multiplying ${x^2}$inside the bracket
$\Rightarrow y = \dfrac{1}{{28}}\left[ {{x^2}\left( {8x - \dfrac{6}{x} + 10} \right)} \right]$
$\Rightarrow y = \dfrac{1}{{28}}\left( {8{x^3} - \dfrac{{6{x^2}}}{x} + 10{x^2}} \right)$

Simplifying the expression, we get,
$y = \dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)$
Now, differentiating $y = \dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)$ with respect to $x$, we have
$\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{28}}\left( {8{x^3} - 6x + 10{x^2}} \right)} \right)$
Taking the constant out of differentiation,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {\dfrac{d}{{dx}}\left( {8{x^3} - 6x + 10{x^2}} \right)} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {\dfrac{d}{{dx}}8{x^3} - \dfrac{d}{{dx}}6x + \dfrac{d}{{dx}}10{x^2}} \right]$

Using power rule of differentiation $\dfrac{d}{{dx}}{x^n} = n \times {x^{n - 1}}$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {8 \times 3 \times {x^{3 - 1}} - 6 \times 1 \times {x^{1 - 1}} + 10 \times 2 \times {x^{2 - 1}}} \right]$
As we know, ${x^0} = 1$. So,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]$
Hence, we have $\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]$
We have to find the value of $\dfrac{{dy}}{{dx}}$at $x = - 1$.
So, we will substitute $x = - 1$in $\dfrac{{dy}}{{dx}} = \dfrac{1}{{28}}\left[ {24{x^2} - 6 + 20x} \right]$

Therefore, $\dfrac{{dy}}{{dx}}$at $x = - 1$is
$\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}\left[ {24{{( - 1)}^2} - 6 + 20( - 1)} \right]$
We know, ${(a)^2} = {( - a)^2} = {a^2}$
$\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}\left[ {24(1) - 6 - 20)} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[24 - 6 - 20]$
Solving the right side
$\dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[24 - 26]$
$\Rightarrow \dfrac{{dy}}{{dx}}{|_{x = - 1}} = \dfrac{1}{{28}}[ - 2]$
Simplifying further, we get
$\therefore \dfrac{{dy}}{{dx}}{|_{x = - 1}} = - \dfrac{1}{{14}}$
Hence, we get, $\dfrac{{dy}}{{dx}}$ at $x = - 1$equals $- \dfrac{1}{{14}}$.

Hence, the correct option is C.

Note: We need to be very careful while finding $f(x)$. We must take care of the fact that we have to replace $x$ by $\dfrac{1}{x}$ in the whole equation and not just the left hand side. And, then keep in mind to solve for $f(x)$ and $f\left( {\dfrac{1}{x}} \right)$ from the obtained and the given equations. Also, we must remember to find the value of $\dfrac{{dy}}{{dx}}$at $x = - 1$ and not just leaving after finding the value of $\dfrac{{dy}}{{dx}}$. We must learn the formulas for differentiation very carefully otherwise we will not be able to solve the question.