Question

y = tan x. x¹⁸ differentiate w.r.t x​

Answer 1

Question :-

$\bf \:Differentiate \: w.r.t.x : \: y \: = {x}^{18} tanx$

Answer

Given : -

$\bf \:y = {x}^{18} tanx$

To find :-

$\bf \:\dfrac{dy}{dx}$

Formula used :-

$\bf \:\dfrac{d}{dx} (u.v) = u\dfrac{d}{dx} v + v\dfrac{d}{dx} u$

$\bf \:\dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}$

$\bf \:\dfrac{d}{dx} tanx = {sec}^{2} x$

Solution :-

$\bf \:Consider \: y = {x}^{18} tanx$

Differentiate w. r. t. x, we get

$\bf \:\dfrac{dy}{dx} = {x}^{18} \dfrac{d}{dx} tanx + tanx\dfrac{d}{dx} {x}^{18}$

$\bf \:\dfrac{dy}{dx} = {x}^{18} {sec}^{2} x + tanx \times 18 {x}^{17}$

$\bf \:\dfrac{dy}{dx} = {x}^{17} (x {sec}^{2} x + 18tanx)$

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