Question

If a function is given by \[f\left( x \right)={{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\text{ then }{{f}^{'}}\left( -\dfrac{1}{2} \right)\]
(a) \[\dfrac{-\sqrt{3}}{2}{{\log }_{e}}\sqrt{3}\]
(b) \[\dfrac{\sqrt{3}}{2}{{\log }_{e}}\sqrt{3}\]
(c) \[-\sqrt{3}{{\log }_{e}}3\]
(d) \[\sqrt{3}{{\log }_{e}}3\]

Answer 1

Hint: In order to solve this question, we will first find the derivative of f(x) and then we will put the value of x as \[\dfrac{-1}{2}\]. To find the derivative of f(x), we should know about a few derivative formulas like chain rule, quotient rule given by - \[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right),\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\text{,}\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a\text{ and }\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]By using these formulas we can solve this question.

Complete step-by-step answer:

In this question, we have been asked to find the value of \[{{f}^{'}}\left( -\dfrac{1}{2} \right)\] where \[f\left( x \right)={{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\]. To solve this question, we should know that the derivatives of \[f\left( g\left( x \right) \right),{{\sin }^{-1}}x,{{a}^{x}},\dfrac{u}{v}\] type function is given by \[\dfrac{d}{dx}\left( f\left( g\left( x \right) \right) \right)={{f}^{'}}\left( g\left( x \right) \right).{{g}^{'}}\left( x \right),\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\text{,}\dfrac{d}{dx}\left( {{a}^{x}} \right)={{a}^{x}}\log a\text{ and }\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]
Now, to solve this question, we will first find the derivative of \[{{\sin }^{-1}}\] then \[\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}}\] and then that of \[{{3}^{x}}\text{ and }{{9}^{x}}\]. So, we can write
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right) \right)\]
Here, we can see that the function is of the form f (g (x)) where \[g\left( x \right)=\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}}\]. So, we can apply the chain rule. Now, we know that \[\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]. So, for \[x=\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}}\], we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{1}{\sqrt{1-{{\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)}^{2}}}}.\dfrac{d}{dx}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\]
Now, we will simplify it. So, we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{1}{\sqrt{\dfrac{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}{{{\left( 1+{{9}^{x}} \right)}^{2}}}}}.\dfrac{d}{dx}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}}}{\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}.\dfrac{d}{dx}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{1+{{9}^{x}}}{\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}.\dfrac{d}{dx}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\]
Now, we know using quotient rule that \[\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}\]. So, we can write \[\dfrac{d}{dx}\left( f\left( x \right) \right)\] for \[u=2\times {{3}^{x}}\text{ and }v=1+{{9}^{x}}\], we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{9}^{x}} \right)}{\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}.\dfrac{\left( 1+{{9}^{x}} \right)\dfrac{d}{dx}\left( 2\times {{3}^{x}} \right)-\left( 2\times {{3}^{x}} \right)\dfrac{d}{dx}\left( 1+{{9}^{x}} \right)}{{{\left( 1+{{9}^{x}} \right)}^{2}}}\]
Now, we know that \[\dfrac{d}{dx}{{a}^{x}}={{a}^{x}}\log a\]. So, we can write \[\dfrac{d}{dx}\left( 2\times {{3}^{x}} \right)=2\times {{3}^{x}}\log 3\] and \[\dfrac{d}{dx}\left( 1+{{9}^{x}} \right)={{9}^{x}}\log 9\]. So, we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{\left( 1+{{9}^{x}} \right)}{\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}.\dfrac{\left( 1+{{9}^{x}} \right)\left( 2\times {{3}^{x}}\times \log 3 \right)-\left( 2\times {{3}^{x}} \right)\left( {{9}^{x}}\log 9 \right)}{{{\left( 1+{{9}^{x}} \right)}^{2}}}\]
Now, we will simplify it further, we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3+{{9}^{x}}\times 2\times {{3}^{x}}\log 3-2\times {{3}^{x}}\times {{9}^{x}}\log 9}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3+{{9}^{x}}\times 2\times {{3}^{x}}\log 3-2\times {{3}^{x}}\times {{9}^{x}}\log {{3}^{2}}}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
Now, we know that \[\log {{a}^{n}}=n\log a\]. So, we can write \[\log {{3}^{2}}=2\log 3\]. So, we get,
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3+{{9}^{x}}\times 2\times {{3}^{x}}\log 3-4\times {{3}^{x}}\times {{9}^{x}}\log 3}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3+\left( {{3}^{x}}\times {{9}^{x}}\log 3 \right)\times \left( 2-4 \right)}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3-2\times {{3}^{x}}\times {{9}^{x}}\log 3}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
\[\dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{2\times {{3}^{x}}\times \log 3\left( 1-{{9}^{x}} \right)}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
We know that \[\dfrac{d}{dx}\left( f\left( x \right) \right)={{f}^{'}}\left( x \right)\]. So, we can write,
\[{{f}^{'}}\left( x \right)=\dfrac{2\times {{3}^{x}}\times \left( 1-{{9}^{x}} \right)\times \log 3}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}\]
Now, we will put \[x=\dfrac{-1}{2}\] to get the value of \[{{f}^{'}}\left( \dfrac{-1}{2} \right)\]. So, we get,
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{2\times {{3}^{\dfrac{-1}{2}}}\times \left( 1-{{9}^{\dfrac{-1}{2}}} \right)\times \log 3}{\left( 1+{{9}^{\dfrac{-1}{2}}} \right)\sqrt{{{\left( 1+{{9}^{\dfrac{-1}{2}}} \right)}^{2}}-{{\left( 2\times {{3}^{\dfrac{-1}{2}}} \right)}^{2}}}}\]
Now, we know that \[\sqrt{9}=3\]. So, we get,
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\dfrac{2}{\sqrt{3}}\left( 1-\dfrac{1}{3} \right)\log 3}{\left( 1+\dfrac{1}{3} \right)\sqrt{{{\left( 1+\dfrac{1}{3} \right)}^{2}}-\dfrac{4}{3}}}\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\log 3}{3\sqrt{3}\left[ \dfrac{4}{3}\sqrt{\dfrac{16}{9}-\dfrac{4}{3}} \right]}\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\log 3}{3\sqrt{3}\left[ \dfrac{4}{3}\sqrt{\dfrac{16-12}{9}} \right]}\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\log 3}{3\sqrt{3}\times \dfrac{4}{3}\times \dfrac{2}{3}}\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\times 3\times 3\times \log 3}{4\times 2\times 3\sqrt{3}}\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\log 3\]
Now, we know that \[3={{\left( \sqrt{3} \right)}^{2}}\]. So, we can write
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\log {{\left( \sqrt{3} \right)}^{2}}\]
And we know that \[\log {{m}^{n}}=n\log m\]. So, we get,
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\left( 2\log \sqrt{3} \right)\]
\[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\sqrt{3}\log \sqrt{3}\]
Hence, we can say that, for \[f\left( x \right)={{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right)\], we get \[{{f}^{'}}\left( \dfrac{-1}{2} \right)=\sqrt{3}{{\log }_{e}}\sqrt{3}\]
Therefore, option (d) is the right answer.

Note: While solving this question, there are high chances of calculation mistakes because it contains a lot of calculations. Also, we need to remember that derivative of \[{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]. Sometimes in a hurry, we end up writing it as \[\dfrac{1}{\sqrt{{{x}^{2}}-1}}\text{ or }\dfrac{-1}{\sqrt{1-{{x}^{2}}}}\] which is wrong. So, we have to remember a few standard derivatives and we have to be very careful while solving the question.