Suppose for a differentiable function h, h(0)=0, h(1)=1 and h′(0)=h′(1)=2. If g(x)=h(eˣ) eʰ⁽ˣ⁾, then g′(0) is equal to:

Solution

Maths sir · 4 steps

Product rule outside, chain rule inside each factor, then put in x = 0.

1. Product rule. Write g(x) = u · v with u = h(eˣ) and v = eʰ⁽ˣ⁾, so g'(x) = u'v + uv'.
2. Chain rule twice. u' = h'(eˣ) · eˣ, since the inner function is eˣ. v' = eʰ⁽ˣ⁾ · h'(x), since the inner function is h(x).

   g'(x) = h'(eˣ) eˣ eʰ⁽ˣ⁾ + h(eˣ) eʰ⁽ˣ⁾ h'(x)

3. Evaluate at zero. At x = 0, e⁰ = 1 and h(0) = 0, so eʰ⁽⁰⁾ = 1. The first term is h'(1) × 1 × 1 = 2. The second term is h(1) × 1 × h'(0) = 1 × 2 = 2.
4. So. g'(0) = 2 + 2 = 4, which is option (a).

The idea to keep. For a product of composite functions, use the product rule first and give each factor its own chain-rule factor. Substitute values only after the full derivative is written.

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