Differentiate the following w.r.t. x.
(i) cos−1(sinx)
(ii) tan−1(1+cosxsinx)
(iii) sin−1(1+4x2x+1)
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Step-by-Step Solution
Step 1: Differentiate (i) cos−1(sinx)
To differentiate the first expression, we use the identity sinx=cos(2π−x). This allows us to simplify the inverse cosine function. Since cos−1(cosθ)=θ, the expression becomes 2π−x. Differentiating this with respect to x gives us −1.
Step 2: Differentiate (ii) tan−1(1+cosxsinx)
For the second expression, we use the half-angle trigonometric identities: sinx=2sin(2x)cos(2x) and 1+cosx=2cos2(2x). Substituting these into the expression inside the tan−1 function simplifies it to tan(2x). Then, tan−1(tanθ)=θ, so the expression becomes 2x. Differentiating 2x with respect to x yields 21.
Step 3: Differentiate (iii) sin−1(1+4x2x+1)
For the third expression, we first rewrite the term inside the sin−1 function as 1+(2x)22⋅2x. We then use a substitution: let 2x=tanθ. This transforms the expression into sin−1(1+tan2θ2tanθ), which simplifies to sin−1(sin2θ), or simply 2θ. Substituting back, we get 2tan−1(2x). Finally, we differentiate 2tan−1(2x) using the chain rule. The derivative of tan−1(u) is 1+u21dxdu, and the derivative of 2x is 2xlog2. Combining these, we get the final result.