Question

if y =
${e}^{log5x} + {e}^{log7x}$
then dy/dx = ?​

Answer 1

Given :-

$y = {e}^{ log(5x) } + {e}^{ log(7x) }$

To Find :-

$\frac{dy}{dx} = </strong><strong>?</strong><strong>$

$\rule{200}{1}$

Solution :-

$y = {e}^{ log(5x) } + {e}^{ log(7x) } \:$

We know that -

• $\frac{d}{dx} ( {e}^{x} ) = {e}^{x}$

$\frac{dy}{dx} = {e}^{ log(5x) } \times \frac{d}{dx}( log(5x)) + {e}^{ log(7x) } \times \frac{d}{dx} ( log(7x))$

We know that -

• $\frac{d}{dx} log(x) = \frac{1}{x}$

$\frac{dy}{dx} = {e}^{ log(5x) } \times \frac{1}{5x} \times \frac{d}{dx} (5x) + {e}^{ log(7x) } \times \frac{1}{7x} \times \frac{d}{dx} (7x) \\ \\ \frac{dy}{dx} = {e}^{ log(5x) } \times \frac{1}{5x} \times 5 + {e}^{ log(7x) } \times \frac{1}{7x} \times 7 \\ \\ \frac{dy}{dx} = \frac{1}{x} {e}^{ log(5x) } + \frac{1}{x} {e}^{ log(7x) } \\ \\ \frac{dy}{dx} = \frac{1}{x} ( {e}^{ log(5x) } + {e}^{ log(7x) } )$

Or we can write it as -

$\frac{dy}{dx} = \frac{1}{x}\times y$

$\boxed{ \frac{dy}{dx} = \frac{y}{x}}$

$\rule{200}{1}$

Mark as Brainliest !! ✨✨