Question

Differentiate the following with respect to x: y= 2log x + 3x + 10​

Answer 1

Question:

Differentiate the following with respect to x: y= 2log x + 3x + 10

Answer :

y= 2log x + 3x + 10

Differentiating w.r.t x

$\tt \implies \frac{dy}{dx} = \frac{d(2logx + 3x + 10 )}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{d(2logx)}{dx} + \frac{d(3x)}{dx} + \frac{d(10)}{dx}$

$\tt \implies \frac{dy}{dx} = \frac{2}{x} + 3 + 0$

$\tt \implies \frac{dy}{dx} = \frac{2}{x} + 3$

$\tt \therefore \frac{dy}{dx} = \frac{2}{x} + 3$

Formulas of Differentiation :

● $\tt \implies \frac{d({x}^{n})}{dx} = n{x}^{n-1}$

● $\tt \implies \frac{d(logx)}{dx} = \frac{1}{x}$

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

●$\tt \implies \frac{d(sinx)}{dx} = cosx$

● $\tt \implies \frac{d(cosx)}{dx} = -sinx$

● $\tt \implies \frac{d(tanx)}{dx} = {sec}^{2}x$

● $\tt \implies \frac{d(cotx)}{dx} = -{cosec}^{2}x$

● $\tt \implies \frac{d(secx)}{dx} = secx.tanx$

● $\tt \implies \frac{d(cosecx)}{dx} = - cosecx.cotx$

Answer 2

Answer:-

You want to know the differentiation of

Y = e^(2logx + 3x)

Taking log both sides

Log (y) = loge^ (2logx + 3x)

Log (y) = (2logx + 3x) log e

Log (y) = 2logx + 3x

Differentiating both sides with respect to x

(1 / y) (dy/dx) = 2 / x + 3

dy/dx = (2/x + 3) y

dy/dx = (2/x + 3)(2logx +3x)

Hope it helps :)