If fx=sin^(-1)(2×3^x)/(1+9^x), then f^(')-1/2 is equal to

Correct answer: (A) √3 log_e√3. Given fx=sin^(-1)(2×3^x)/(1+9^x) Let 3^x=tant⇒t=tan^(-1)3^x So, f(x)=sin^(-1)(2tan⁡t)/(1+tan^2t) We know that, sin(2t)=(2tan(t))/(1+tan^2(t)) ⇒f(x)=sin^(-1)sin2t ∴f(x)=2t=2tan^(-1)3^x ⇒dfx/dx=2/(1+(3x)^2 )×3^x.log_e3 At x=1/2, df/dx=2/(1+312^2)×3^12.log_e3 =1/2×√3×log_e3 =√3×log_e√3

Q42 / 46 · NTA Abhyas (JEE Main) 2020 · Mathematics · Differentiation

If $f (x) = {s i n}^{- 1} (\frac{2 \times 3^{x}}{1 + 9^{x}}),$ then $f^{'} (- \frac{1}{2})$ is equal to

Correct answer
A
$\sqrt{3} {l o g}_{e} \sqrt{3}$
B
$- \sqrt{3} {l o g}_{e} \sqrt{3}$
C
$- \sqrt{3} {l o g}_{e} 3$
D
$\sqrt{3} {l o g}_{e} 3$

Solution

Given

f (x) = s i n^{- 1} (\frac{2 \times 3^{x}}{1 + 9^{x}})

Let

3^{x} = t a n (t) \Rightarrow t = \tan^{- 1} (3^{x})

So,

f (x) = s i n^{- 1} (\frac{2 \tan (t)}{1 + t a n^{2} (t)})

We know that,

s i n (2 t) = \frac{2 t a n (t)}{1 + {t a n}^{2} (t)}

\Rightarrow f (x) = s i n^{- 1} (s i n (2 t))

∴ f (x) = 2 t = 2 t a n^{- 1} (3^{x})

\Rightarrow \frac{d f (x)}{d x} = \frac{2}{1 + {(3^{x})}^{2}} \times 3^{x} . l o g_{e} 3

x = \frac{1}{2}, \frac{d f}{d x} = \frac{2}{1 + {(3^{\frac{1}{2}})}^{2}} \times 3^{\frac{1}{2}} . l o g_{e} 3

= \frac{1}{2} \times \sqrt{3} \times {l o g}_{e} 3

= \sqrt{3} \times {l o g}_{e} \sqrt{3}

Hence, the correct answer is option: 'A'.

If fx=sin-12×3x1+9x, then f'-12 is equal to

Solution

More Differentiation Questions

If $f (x) = {s i n}^{- 1} (\frac{2 \times 3^{x}}{1 + 9^{x}}),$ then $f^{'} (- \frac{1}{2})$ is equal to