Question

Differentiate x^{-ln x} with respect to \(\rm e^{x^{2}}\)
1. \(\rm -x^{-\ln x}(\ln x)\over x^2e^{x^{2}}\)
2. \(\rm -x^{-\ln x}(\ln x)\over xe^{x^{2}}\)
3. \(\rm -2x^{-\ln x}(\ln x)\over x^2e^{x^{2}}\)
4. \(\rm -2x^{-\ln x}(\ln x)\over xe^{x^{2}}\)

Answer 1

Correct Answer - Option 1 : \(\rm -x^{-\ln x}(\ln x)\over x^2e^{x^{2}}\)

Concept:

Parametric Form:

If f(x) and g(x) are the functions in x, then 

\(\rm df(x)\over dg(x)\) = \(\rm \frac{df(x)\over dx}{dg(x)\over dx}\) 

Calculation:

Let z = x^{-ln x} 

Taking log both sides, we get

ln z = ln x-ln x

ln z = - (ln x)²                       (∵ ln mⁿ = n ln m)

Differentiating with respect to x

\(\rm {1\over z}{dz\over dx}\) = -2 ln x\(\rm\left({1\over x}\right)\)

\(\rm {dz\over dx} = z\left[-2\ln x\over x\right]\)

\(\rm {dz\over dx} = x^{-\ln x}\left[-2\ln x\over x\right]\)

\(\rm {dz\over dx} = -2x^{-\ln x}\left[\ln x\over x\right]\)

Also y = \(\rm e^{x^{2}}\)

Taking log both sides, we get

ln y = ln \(\rm e^{x^{2}}\)        

ln y = x² ln e               (∵ ln mn = n ln m)

ln y = x²                      (∵ ln e = 1)

Differentiating with respect to x

\(\rm {1\over y}{dy\over dx} = 2x\)

\(\rm {dy\over dx} = y[2x]\)

\(\rm {dy\over dx} = 2x e^{x^{2}}\)

The required result is

\(\rm dz\over dy\) = \(\rm {dz\over dx}\over{dy\over dx}\)

= \(\rm -2x^{-\ln x}\left[\ln x\over x\right]\over2xe^{x^{2}}\)

= \(\rm -x^{-\ln x}(\ln x)\over x^2e^{x^{2}}\)