%%

Suppose for a differentiable function $$h, h(0)=0, h(1)=1$$ and $$h^{\prime}(0)=h^{\prime}(1)=2$$. If $$g(x)=h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$, then $$g^{\prime}(0)$$ is equal to:

JEE · Math · previous-year question

  1. A.4correct
  2. B.5
  3. C.3
  4. D.8

Answer

A. 4

Explanation

To determine $$g^{\prime}(0)$$, we start by applying the chain rule and product rule to find the derivative of the given function $$g(x) = h\left(\mathrm{e}^x\right) \mathrm{e}^{h(x)}$$. The product rule states that if we have two functions $$u(x)$$ and $$v(x)$$, then the derivative of their product is given by: $$ (u(x)v(x))' = u'(x)v(x) + u(x)v'(x) $$ Let's denote $$u(x) = h(\mathrm{e}^x)$$ and $$v(x) = \mathrm{e}^{h(x)}$$. First, we need to find $$u'(x)$$ and $$v'(x)$$. Using the chain rule, we find: $$ u'(x) = \frac{d}{dx}h(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \frac{d}{dx}(\mathrm{e}^x) = h'(\mathrm{e}^x) \cdot \mathrm{e}^x $$ Now, the derivative of $$v(x)$$ is: $$ v'(x) = \frac{d}{dx}(\mathrm{e}^{h(x)}) = \mathrm{e}^{h(x)} \cdot h'(x) $$ Using the product rule, we get the derivative of $$g(x)$$: $$ g'(x) = u'(x) v(x) + u(x) v'(x) $$ Substituting $$u(x)$$, $$v(x)$$, $$u'(x)$$, and $$v'(x)$$ into the above expression, we get: $$ g'(x) = \left( h'(\mathrm{e}^x) \mathrm{e}^x \right) \mathrm{e}^{h(x)} + \left( h(\mathrm{e}^x) \right) \left( \mathrm{e}^{h(x)} h'(x) \right) $$ Next, we need to evaluate this at $$x = 0$$: First, we know that: $$ h(0) = 0 $$ $$ h(1) = 1 $$ $$ h'(0) = 2 $$ $$ h'(1) = 2 $$ Substituting $$x = 0$$ into the expressions, we get: $$ u(0) = h(\mathrm{e}^0) = h(1) = 1 $$ $$ v(0) = \mathrm{e}^{h(0)} = \mathrm{e}^0 = 1 $$ $$ u'(0) = h'(\mathrm{e}^0) \mathrm{e}^0 = h'(1) \cdot 1 = 2 $$ $$ v'(0) = \mathrm{e}^{h(0)} h'(0) = \mathrm{e}^0 \cdot 2 = 2 $$ Therefore, evaluating $$g'(0)$$: $$ g'(0) = \left( u'(0) v(0) \right) + \left( u(0) v'(0) \right) $$ $$ g'(0) = \left( 2 \cdot 1 \right) + \left( 1 \cdot 2 \right) $$ $$ g'(0) = 2 + 2 = 4 $$ Thus, the value of $$g^{\prime}(0)$$ is 4, which corresponds to Option A. The correct answer is Option A: 4.

Practice more JEE questions

Answer thousands more real JEE previous-year questions free, get graded instantly, and climb the global ranked leaderboard.

Practice JEE free →

More JEE Math questions