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Differentiate the following w.r.t. x : cos-1 (1-9x)(1+9x)

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प्रश्न

Differentiate the following w.r.t. x :

`cos^-1  ((1 - 9^x))/((1 + 9^x)`

योग
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उत्तर

Let y = `cos^-1  ((1 - 9^x))/((1 + 9^x)`

= `cos^-1[(1 - (3^x)^2)/(1 + (3^x)^2)]`
Put 3x = tanθ.
Then θ = tan–1(3x)

∴ y = `cos^-1((1 - tan^2θ)/(1 + tan^2θ))`
= cos–1(cos2θ)
= 2θ
= 2tan–1(3x)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[2tan^-1(3^x)]`

= `2"d"/"dx"[tan^-1 (3^x)]`

= `2 xx (1)/(1 + (3^x)^2)."d"/"dx"(3^x)`

= `(2)/(1 + 3^(2x)) xx 3^xlog3`

= `(2.3^xlog3)/(1 + 3^(2x)`

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अध्याय 1: Differentiation - Exercise 1.2 [पृष्ठ ३०]

APPEARS IN

बालभारती Mathematics and Statistics 2 (Arts and Science) [English] Standard 12 Maharashtra State Board
अध्याय 1 Differentiation
Exercise 1.2 | Q 9.07 | पृष्ठ ३०

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