Let u = cos–1(4x3 – 3x) and v = tan-1\(\Big(\cfrac{\sqrt{1-\text x^2}}{\text x}\Big)\)
We need to differentiate u with respect to v that is find \(\cfrac{du}{dv}.
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We have u = cos–1(4x3 – 3x)
By substituting x = cos θ,
we have u = cos–1(4cos3θ – 3cosθ)
But, cos3θ = 4cos 3θ – 3cosθ
⇒ u = cos –1(cos3θ)
Hence, u = cos–1(cos3θ) = 3θ
⇒ u = 3cos–1x
On differentiating u with respect to x, we get
\(\cfrac{du}{d\mathrm x}=\cfrac{d}{d\text x}
\)(3 cos-1x)
Hence, v = tan–1(tanθ) = θ
⇒ v = cos–1x
On differentiating v with respect to x, we get
We have
