Question

Find $\dfrac{{dy}}{{dx}}$ of $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$.

Answer 1

Hint : In the given problem, we are required to differentiate the given function $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ with respect to x. Since, $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ is a composite function, so we will have to apply chain rule of differentiation in the process of differentiating $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ . So, differentiation of $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ with respect to x will be done layer by layer using the chain rule of differentiation. Also derivatives of ${\sin ^{ - 1}}\left( x \right)$ and ${a^x}$ with respect to $x$ must be remembered in order to solve the given problem.

Complete step-by-step answer :
So, we have to find the derivative of $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ with respect to $x$ using the chain rule of differentiation.
Now, $\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)} \right]$
Now, Let us assume $u = \left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$. So substituting $\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$as $u$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left[ {{{\sin }^{ - 1}}\left( u \right)} \right]$
Now, we know that the derivative of ${\sin ^{ - 1}}\left( x \right)$ with respect to x is $\dfrac{1}{{\sqrt {1 - {x^2}} }}$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {u^2}} }}\dfrac{{du}}{{dx}}$
Now, putting back $u$as $\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\dfrac{d}{{dx}}\left[ {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right]$
Now, we have to differentiate $\left[ {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right]$ with respect to x using the quotient rule of differentiation. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\left( {1 + {4^x}} \right)\dfrac{{d\left( {{2^{x + 1}}} \right)}}{{dx}} - {2^{x + 1}}\dfrac{{d\left( {1 + {4^x}} \right)}}{{dx}}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Now, we know that the derivative of ${a^x}$ with respect to x is ${a^x}\log a$. So, simplifying the expression, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\left( {1 + {4^x}} \right)\left( {{2^{x + 1}}} \right)\left( {\log 2} \right) - \left( {{2^{x + 1}}} \right)\left( {{4^x}} \right)\left( {\log 4} \right)}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Taking $\left( {{2^{x + 1}}} \right)$ common from the bracket, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)}}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\left( {1 + {4^x}} \right)\left( {\log 2} \right) - \left( {{4^x}} \right)\left( {\log 4} \right)}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Also, we know that $\log 4 = \log {2^2} = 2\log 2$. Hence, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)}}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\left( {1 + {4^x}} \right)\left( {\log 2} \right) - 2\left( {{4^x}} \right)\left( {\log 2} \right)}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Opening brackets and simplifying the calculations, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)}}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\log 2 + {4^x}\log 2 - 2\left( {{4^x}\log 2} \right)}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)}}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{\log 2 - {4^x}\log 2}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Taking $\left( {\log 2} \right)$ common from the bracket, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\log 2}}{{\sqrt {1 - {{\left( {\;\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)}^2}} }}\left[ {\dfrac{{1 - {4^x}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\log 2}}{{\sqrt {\dfrac{{{{\left( {1 + {4^x}} \right)}^2} - {{\left( {{2^{x + 1}}} \right)}^2}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} }}\left[ {\dfrac{{1 - {4^x}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Evaluating the square root, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\log 2}}{{\sqrt {\dfrac{{{{\left( {1 + {4^x}} \right)}^2} - {{\left( {{2^{x + 1}}} \right)}^2}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} }}\left[ {\dfrac{{1 - {4^x}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)\left( {1 + {4^x}} \right)}}{{\sqrt {{{\left( {1 + {4^x}} \right)}^2} - {{\left( {{2^{x + 1}}} \right)}^2}} }}\left[ {\dfrac{{1 - {4^x}}}{{{{\left( {1 + {4^x}} \right)}^2}}}} \right]$
Cancelling out the common factor in numerator and denominator, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)}}{{\sqrt {{{\left( {1 + {4^x}} \right)}^2} - {{\left( {{2^{x + 1}}} \right)}^2}} }}\left[ {\dfrac{{1 - {4^x}}}{{1 + {4^x}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)}}{{\sqrt {\left( {{4^{2x}} + 2\left( {{4^x}} \right) + 1} \right) - 4{{\left( {{2^x}} \right)}^2}} }}\left[ {\dfrac{{1 - {4^x}}}{{1 + {4^x}}}} \right]$
Now, we know that ${\left( {{2^x}} \right)^2} = {2^{2x}} = {4^x}$. So, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)}}{{\sqrt {\left( {{4^{2x}} + 2\left( {{4^x}} \right) + 1} \right) - 4\left( {{4^x}} \right)} }}\left[ {\dfrac{{1 - {4^x}}}{{1 + {4^x}}}} \right]$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)}}{{\sqrt {\left( {{4^{2x}} - 2\left( {{4^x}} \right) + 1} \right)} }}\left[ {\dfrac{{1 - {4^x}}}{{1 + {4^x}}}} \right]$
We know that ${\left( {1 - {4^x}} \right)^2} = {4^{2x}} - 2\left( {{4^x}} \right) + 1$. So, we get, $\sqrt {\left( {{4^{2x}} - 2\left( {{4^x}} \right) + 1} \right)} = \left( {1 - {4^x}} \right)$
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {{2^{x + 1}}} \right)\left( {\log 2} \right)}}{{\left( {1 - {4^x}} \right)}}\left[ {\dfrac{{1 - {4^x}}}{{1 + {4^x}}}} \right]$
Cancelling the common factors in numerator and denominator, we get,
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{2^{x + 1}}\log 2}}{{1 + {4^x}}}$
So, the value of $\dfrac{{dy}}{{dx}}$ if $y = {\sin ^{ - 1}}\left( {\dfrac{{{2^{x + 1}}}}{{1 + {4^x}}}} \right)$ is $\left( {\dfrac{{{2^{x + 1}}\log 2}}{{1 + {4^x}}}} \right)$

Note : The given problem may also be solved using the first principle of differentiation. The derivatives of basic functions must be learned by heart in order to find derivatives of complex composite functions using chain rule of differentiation. 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.