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Question: If a function is given by \[f\left( x \right)={{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^...

If a function is given by f(x)=sin⁡−1(2×3x1+9x) then f′(−12)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) −32log⁡e3\dfrac{-\sqrt{3}}{2}{{\log }_{e}}\sqrt{3}
(b) 32log⁡e3\dfrac{\sqrt{3}}{2}{{\log }_{e}}\sqrt{3}
(c) −3log⁡e3-\sqrt{3}{{\log }_{e}}3
(d) 3log⁡e3\sqrt{3}{{\log }_{e}}3

Explanation

Solution

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 −12\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 - ddx(f(g(x)))=f′(g(x)).g′(x),ddxsin⁡−1x=11−x2,ddx(ax)=axlog⁡a and ddx(uv)=vdudx−udvdxv2\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′(−12){{f}^{'}}\left( -\dfrac{1}{2} \right) where f(x)=sin⁡−1(2×3x1+9x)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(g(x)),sin⁡−1x,ax,uvf\left( g\left( x \right) \right),{{\sin }^{-1}}x,{{a}^{x}},\dfrac{u}{v} type function is given by ddx(f(g(x)))=f′(g(x)).g′(x),ddxsin⁡−1x=11−x2,ddx(ax)=axlog⁡a and ddx(uv)=vdudx−udvdxv2\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{{\sin }^{-1}} then 2×3x1+9x\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} and then that of 3x and 9x{{3}^{x}}\text{ and }{{9}^{x}}. So, we can write
ddx(f(x))=ddx(sin⁡−1(2×3x1+9x))\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(x)=2×3x1+9xg\left( x \right)=\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}}. So, we can apply the chain rule. Now, we know that ddxsin⁡−1x=11−x2\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}. So, for x=2×3x1+9xx=\dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}}, we get,
ddx(f(x))=11−(2×3x1+9x)2.ddx(2×3x1+9x)\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,
ddx(f(x))=1(1+9x)2−(2×3x)2(1+9x)2.ddx(2×3x1+9x)\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)
ddx(f(x))=(1+9x)2(1+9x)2−(2×3x)2.ddx(2×3x1+9x)\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)
ddx(f(x))=1+9x(1+9x)2−(2×3x)2.ddx(2×3x1+9x)\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 ddx(uv)=vdudx−udvdxv2\dfrac{d}{dx}\left( \dfrac{u}{v} \right)=\dfrac{v\dfrac{du}{dx}-u\dfrac{dv}{dx}}{{{v}^{2}}}. So, we can write ddx(f(x))\dfrac{d}{dx}\left( f\left( x \right) \right) for u=2×3x and v=1+9xu=2\times {{3}^{x}}\text{ and }v=1+{{9}^{x}}, we get,
ddx(f(x))=(1+9x)(1+9x)2−(2×3x)2.(1+9x)ddx(2×3x)−(2×3x)ddx(1+9x)(1+9x)2\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 ddxax=axlog⁡a\dfrac{d}{dx}{{a}^{x}}={{a}^{x}}\log a. So, we can write ddx(2×3x)=2×3xlog⁡3\dfrac{d}{dx}\left( 2\times {{3}^{x}} \right)=2\times {{3}^{x}}\log 3 and ddx(1+9x)=9xlog⁡9\dfrac{d}{dx}\left( 1+{{9}^{x}} \right)={{9}^{x}}\log 9. So, we get,
ddx(f(x))=(1+9x)(1+9x)2−(2×3x)2.(1+9x)(2×3x×log⁡3)−(2×3x)(9xlog⁡9)(1+9x)2\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,
ddx(f(x))=2×3x×log⁡3+9x×2×3xlog⁡3−2×3x×9xlog⁡9(1+9x)(1+9x)2−(2×3x)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 9}{\left( 1+{{9}^{x}} \right)\sqrt{{{\left( 1+{{9}^{x}} \right)}^{2}}-{{\left( 2\times {{3}^{x}} \right)}^{2}}}}
ddx(f(x))=2×3x×log⁡3+9x×2×3xlog⁡3−2×3x×9xlog⁡32(1+9x)(1+9x)2−(2×3x)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⁡an=nlog⁡a\log {{a}^{n}}=n\log a. So, we can write log⁡32=2log⁡3\log {{3}^{2}}=2\log 3. So, we get,
ddx(f(x))=2×3x×log⁡3+9x×2×3xlog⁡3−4×3x×9xlog⁡3(1+9x)(1+9x)2−(2×3x)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 3-4\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}}}}
ddx(f(x))=2×3x×log⁡3+(3x×9xlog⁡3)×(2−4)(1+9x)(1+9x)2−(2×3x)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}}}}
ddx(f(x))=2×3x×log⁡3−2×3x×9xlog⁡3(1+9x)(1+9x)2−(2×3x)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}}}}
ddx(f(x))=2×3x×log⁡3(1−9x)(1+9x)(1+9x)2−(2×3x)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 ddx(f(x))=f′(x)\dfrac{d}{dx}\left( f\left( x \right) \right)={{f}^{'}}\left( x \right). So, we can write,
f′(x)=2×3x×(1−9x)×log⁡3(1+9x)(1+9x)2−(2×3x)2{{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=−12x=\dfrac{-1}{2} to get the value of f′(−12){{f}^{'}}\left( \dfrac{-1}{2} \right). So, we get,
f′(−12)=2×3−12×(1−9−12)×log⁡3(1+9−12)(1+9−12)2−(2×3−12)2{{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 9=3\sqrt{9}=3. So, we get,
f′(−12)=23(1−13)log⁡3(1+13)(1+13)2−43{{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′(−12)=4log⁡333[43169−43]{{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′(−12)=4log⁡333[4316−129]{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\log 3}{3\sqrt{3}\left[ \dfrac{4}{3}\sqrt{\dfrac{16-12}{9}} \right]}
f′(−12)=4log⁡333×43×23{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\log 3}{3\sqrt{3}\times \dfrac{4}{3}\times \dfrac{2}{3}}
f′(−12)=4×3×3×log⁡34×2×33{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{4\times 3\times 3\times \log 3}{4\times 2\times 3\sqrt{3}}
f′(−12)=32log⁡3{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\log 3
Now, we know that 3=(3)23={{\left( \sqrt{3} \right)}^{2}}. So, we can write
f′(−12)=32log⁡(3)2{{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\log {{\left( \sqrt{3} \right)}^{2}}
And we know that log⁡mn=nlog⁡m\log {{m}^{n}}=n\log m. So, we get,
f′(−12)=32(2log⁡3){{f}^{'}}\left( \dfrac{-1}{2} \right)=\dfrac{\sqrt{3}}{2}\left( 2\log \sqrt{3} \right)
f′(−12)=3log⁡3{{f}^{'}}\left( \dfrac{-1}{2} \right)=\sqrt{3}\log \sqrt{3}
Hence, we can say that, for f(x)=sin⁡−1(2×3x1+9x)f\left( x \right)={{\sin }^{-1}}\left( \dfrac{2\times {{3}^{x}}}{1+{{9}^{x}}} \right), we get f′(−12)=3log⁡e3{{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⁡−1x=11−x2{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}. Sometimes in a hurry, we end up writing it as 1x2−1 or −11−x2\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.