Question

Find the
derivative
of y=e^x+log x +3x^2-Cos x​

Answer 1

Answer:

$\frac{dy}{dx} = e^x + 1/x + 6x + sinx$

Step-by-step explanation:

y=e^x + log x + 3x² - cos x​

$\frac{dy}{dx} = \frac{d}{dx}[e^x + log x + 3x^2 - cos x] = \frac{d}{dx}(e^x) + \frac{d}{dx} (logx) + \frac{d}{dx} (3x^2) + \frac{d}{dx} (-cos x)$

differentiate individually

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

$\frac{d}{dx} (log x) = 1/x$

$\frac{d}{dx}(3x^2) = 3\frac{d}{dx}(x^2) = 3 * 2x = 6x$

$\frac{d}{dx}(- cos x) = -\frac{d}{dx}(cos x) = -(-sin x) = sin x$

apply these values in the equation

we get

$\frac{dy}{dx} = e^x + 1/x + 6x + sinx$