Let f be a differentiable function in (0, 2) such that f (x) = \int_{0}^{6} \textrm{ } \left(\left(\right. \frac{6 - t}{t} \left.\right)\right)^{x \textrm{ } - \textrm{ } 1} \frac{dt}{6 - t} then f \textrm{ } ' \textrm{ } \left(\right. \frac{3}{2} \left.\right) is equal to

Engineering

Mathematics

Leibnitz Rule Derivative of Anti Derivatives

Question

Let f be a differentiable function in (0, 2) such that f (x) = $\int_{0}^{6} {(\frac{6 - t}{t})}^{x - 1} \frac{dt}{6 - t}$ then $f' (\frac{3}{2})$ is equal to

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f (x) = $\int_{0}^{6} {(\frac{6 - t}{t})}^{x - 1} \frac{dt}{6 - t}$    .....(1)
Applying, king property
f (x) = $\int_{0}^{6} {(\frac{t}{6 - t})}^{x - 1} \frac{dt}{t}$
Replacing x by – x + 3
f (– x + 3) = $\int_{0}^{6} {(\frac{t}{6 - t})}^{- x + 2} \frac{dt}{t}$
f (– x + 3) = $\int_{0}^{6} {(\frac{6 - t}{t})}^{x - 2} \frac{dt}{t}$    ....(2)
From equation (1) and (2), we get
f (x) = f (3 – x)
⇒   f ' (x) = – f ' (3 – x)   ⇒ $f' (\frac{3}{2})$ = 0