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प्रश्न
Differentiate the following w.r.t. x : `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
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उत्तर
Let y = `cos^-1((e^x - e^(-x))/(e^x + e^(-x)))`
= `cos^-1[(e^x - 1/e^x)/(e^x + 1/e^x)]`
= `cos^-1((e^(2x) - 1)/(e^(2x) + 1))`
Put ex = tanθ.
Then θ = tan–1(ex)
∴ y = `cos^-1((tan^2θ - 1)/(tan^2θ + 1))`
= `cos^-1[-((1 - tan^2θ)/(1 + tan^2θ))]`
= cos–1(– cos2θ)
= cos–1[cos(π – 2θ)]
= π – 2θ
= π – 2tan–1(ex)
Differentiating w.r.t. x, we get
`"dy"/"dx" = "d"/"dx"[pi - 2tan^-1(e^x)]`
= `"d"/"dx"(pi) - 2"d"/"dx"[tan^-1(e^x)]`
= `0 - 2 xx (1)/(1 + (e^x)^2)."d"/"dx"(e^x)`
= `(-2)/(1 + e^(2x)) xx e^x`
= `-(2e^x)/(1 + e^(2x)`
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