Question

y = 2x.sinx + 3 cos x - 4x.ln x​

Answer 1

CORRECT QUESTION:

Differentiate y = 2x.sinx + 3 cos x - 4x.ln x

ANSWER:

$y' = 2 x\cos x - \sin x - 4 - 4\:ln(x)$

GIVEN:

y = 2x.sinx + 3 cos x - 4x.ln x

TO FIND:

dy/dx or y'

FORMUALE:

$\dfrac{d}{dx} (u.v) = uv' + vu'\\ \\ \\\dfrac{d}{dx} \sin x = \cos x\\ \\ \\ \dfrac{dx}{dx} = 1\\ \\ \\ \dfrac{d}{dx} \cos x = - \sin x\\ \\ \\ \dfrac{d}{dx} ln(x) = \dfrac{1}{x}$

EXPLANATION:

Let y = g(x) + h(x) + i(x)

Also y' = g'(x) + h'(x) + i'(x)

Let g(x) = 2x.sinx

Differentiate with respect to x

$g'(x) = 2x \left( \dfrac{d}{dx} \sin x \right) + 2 \sin x \left( \dfrac{dx}{dx}\right) \\ \\ \\ g'(x) = 2x \cos x\bigg(\frac{dx}{dx}\bigg) + 2 \sin x \\ \\ \\g'(x) = 2 \sin x + 2x \cos x$

Let h(x) = 3 cos x

Differentiate with respect to x

$h'(x) =3 \left( \dfrac{d}{dx} \cos x \right) \\ \\ \\ h'(x)= - 3\sin x\bigg(\dfrac{dx}{dx}\bigg)\\ \\ \\h'(x)= - 3\sin x$

Let i(x) = - 4x.ln(x)

Differentiate with respect to x

$i'(x)= - 4x \bigg(\dfrac{d}{dx} ln(x) \bigg) - 4 ln(x) \bigg( \frac{dx}{dx} \bigg)\\ \\ \\ i'(x)= -4\bigg(\dfrac{x}{x}\bigg)\bigg(\dfrac{dx}{dx}\bigg)\\ \\ \\i'(x) = - 4 - 4\:ln(x)$

$g'(x) + h'(x) + i'(x) = y' \\ \\ \\ y' = 2\sin x + 2 x\cos x - 3\sin x - 4 - 4\:ln(x) \\ \\ \\ y' = 2x \cos x - \sin x - 4 - 4\:ln(x)$