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Differentiate the following w.r.t. x : cos-1[3cos(ex)+2sin(ex)13]

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प्रश्न

Differentiate the following w.r.t. x :

`cos^-1[(3cos(e^x) + 2sin(e^x))/sqrt(13)]`

योग
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उत्तर

y = `cos^-1((3cos(e^x) + 2sin(e^x))/sqrt(13))`

y  = `cos^-1(cos(e^x).(3)/sqrt(13) + sin(e^x)2/sqrt(13))`

Put, `(3)/sqrt(13) = cos α, (2)/sqrt(13) = sin α`

Also,

sin2α + cos2α = `(9)/(13) + (4)/(13)` = 1

And,

tan α = `(2)/(3)       ∴ α = "tan"^-1(2/3)`

y = cos–1(cos ex. cos α + sin ex. sin α)

y = cos–1[cos(ex – α)]       ...[∵ cos–1x.(cos x) = x]

y = ex – α

y = `e^x  –  tan^-1(2/3)`

Differentiating w.r.t. x, we get

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

`dy/dx` = ex – 0

`dy/dx` = ex

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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 8.5 | पृष्ठ ३०

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