Suppose f(x)=((2ˣ+2⁻ˣ) tan x √(tan ⁻¹(x²-x+1)))/(7 x²+3 x+1)³. Then the value of f′(0) is equal to

Solution

Maths sir · 4 steps

Pull out the tan x factor: it is 0 at x = 0, so only its derivative survives in the product rule.

1. Split the function. Write f(x) = tan x · g(x), where g(x) = (2ˣ+2⁻ˣ)√tan⁻¹(x²-x+1)/(7x²+3x+1)³. Everything except tan x goes into g.

   f'(0) = sec² 0 · g(0) + tan 0 · g'(0) = 1 · g(0) + 0

2. No need for g′. Since tan 0 = 0, the term with g'(0) vanishes. You only need the value g(0), not its messy derivative.
3. Evaluate g(0). At x = 0: 2⁰ + 2⁰ = 2, tan⁻¹(1) = π/4 so its square root is √π/2, and the denominator is 1³ = 1. So g(0) = 2 × √π/2 = √π.
4. So. f'(0) = g(0) = √π, which is option (b).

The idea to keep. If a product has a factor that is zero at the point, its derivative there is just (derivative of that factor) times (the rest evaluated at the point). Never differentiate the whole mess.

Answer from the NTA final answer key. Solution by PrepWiser.