Differentiate the following w.r.t. x. (i) cos^-1(sin x) (ii) tan^-1(sin x / (1 + cos x)) (iii) sin^-1(2^(x+1) / (1 + 4x))
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Step-by-Step Solution
Step 1: Differentiate (i) cos−1(sinx)
To differentiate cos−1(sinx), we use the chain rule. The derivative of cos−1(u) is 1−u2−1⋅dxdu. Here, u=sinx.
Step 2: Simplify (i)
We know that dxd(sinx)=cosx. Also, using the identity sin2x+cos2x=1, we can replace 1−sin2x with cos2x. The square root of cos2x is ∣cosx∣.
Step 3: Differentiate (ii) tan−1(1+cosxsinx)
First, we simplify the argument of the tan−1 function using half-angle identities: sinx=2sin(x/2)cos(x/2) and 1+cosx=2cos2(x/2). This simplifies the expression to tan(x/2).
Step 4: Differentiate simplified (ii)
Since tan−1(tanθ)=θ, the expression simplifies to x/2. Now, we differentiate x/2 with respect to x.
Step 5: Differentiate (iii) sin−1(1+4x2x+1)
Let's rewrite the argument of the sin−1 function. We can express 2x+1 as 2⋅2x and 4x as (2x)2. This transformation prepares the expression for a substitution.
Step 6: Substitute and simplify (iii)
Let 2x=tanθ. Then the expression becomes 1+tan2θ2tanθ. This is a known trigonometric identity for sin(2θ).
Step 7: Differentiate simplified (iii)
Since sin−1(sin(2θ))=2θ, we substitute back θ=tan−1(2x). Now we need to differentiate 2tan−1(2x) with respect to x.
Step 8: Final differentiation of (iii)
The derivative of tan−1(u) is 1+u21⋅dxdu. Here u=2x, and dxd(2x)=2xln2. Combining these, we get the final derivative.