Question

If y = `{f(x)}^{phi(x)}`, then `dy/dx` is ______ 

Options

• `e^(phi log f){phi/f . (df)/dx + logf. (dphi)/dx}`

• `phi/f  ((df)/dx) + (dphi)/dx log f`

• `e^(phi log f) {phi f^'/f + phi^' logf^'}`

• None of these

Answer 1

If y = `{f(x)}^{phi(x)}`, then `dy/dx` is `underline(e^(phi log f){phi/f . (df)/dx + logf. (dphi)/dx})`.

Explanation:

y = `{f(x)}^{phi(x)}`

Taking logarithm on both sides, we get

log y = φ(x) log {f(x)}

⇒ y = `e^{phi(x) logf(x)}`

∴ `dy/dx = e^{phi(x)logf(x)}. d/dx[phi(x) logf(x)]`

= `e^{phi(x)logf(x)} {phi(x). (f^'(x))/(f(x)) + log f(x).phi^'(x)}`