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Question
Differentiate \[\sin^{- 1} \left\{ \frac{2^{x + 1} \cdot 3^x}{1 + \left( 36 \right)^x} \right\}\] with respect to x ?
Solution
We have ,
y = `sin^-1[(2^(x + 1) . 3^x)/(1 + 36^x)]`
y = `sin^-1[(2 . 6^x)/(1 + 6^(2x))]`
put `6^x = tanθ`
⇒ θ = `tan^-1(6^x)`
Now ,
y = `sin^-1[(2 tanθ)/(1 + tan^2θ)]`
y = `sin^-1[sin 2θ]`
y = 2 θ
y = `2tan^-1(6^x)`
`dy/dx = 2/(1 + 6^(2x)) xx d/dx (6^x)`
`dy/dx = 2/(1 + 6^(2x)) xx 6^x log 6`
`dy/dx = 2/(1 + (36)^(x)) xx 6^x log 6`
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