Answer:
[tex]\frac{dy}{dx} = e^{x^{2} } ( \frac{1}{x} + 2x ln2x )
Step-by-step explanation:
Let Given function (y) = [tex]e^{x^{2}} ln2x[/tex]
If we differentiate this function with respect to x -
[tex]\frac{dy}{dx} = \frac{d}{dx} ( e^{x^{2} } ln2x)[/tex]
As we know that-
\frac{d}{dx} ( I × II ) = I × \frac{d}{dx} ( II ) + II × \frac{d}{dx} (I)
[tex]\frac{dy}{dx} = e^{x^{2} } \frac{d}{dx} ( ln2x ) + ln2x \frac{d}{dx} ( e^{x^{2} })
[tex]\frac{dy}{dx} = e^{x^{2} } \frac{1}{2x} × 2 + ln2x × e^{x^{2} } × 2x
[tex]\frac{dy}{dx} = e^{x^{2} } \frac{1}{x} + ln2x × e^{x^{2} } × 2x
[tex]\frac{dy}{dx} = e^{x^{2} } ( \frac{1}{x} + 2x ln2x )
