Question

solve dy/dx if y= sin(1-2x)^3

Answer 1

Question-To differenitaiate y=sin(1-2x)^3 with respect to x

Solution:-

y=(sin(1-2x)^3

We will use chain rule to differenitaiate it

dy/dx=dy/dx {sin(1-2x)^3} ×dy/dx sin(1-2x)×dy/dx(1-2x)

=>3sin^2(1-2x)×cos(1-2x)×-2

=>-6sin^2(1-2x)cos(1-2x)

simplifying more we get

=>-3×sin(1-2x)×2sin(1-2x)cos(1-2x)

=>-3sin(1-2x)×sin2(1-2x)

=>-3sin(1-2x)sin2(1-2x)

Hence the differentiation of y=sin(1-2x)^3 is -3sin(1-2x)sin2(1-2x)

{hope it helps you}

Answer 2

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = -6.\sin^2\left(1-2x\right).\left(\cos\left(1-2x\right)\right)$

Explanation:

Given :

$\displaystyle \sf y = \sin\left(1-2x\right)^3$

We're asked to find d y / d x

Using chain rule here :

Diff. w.r.t. x :

$\displaystyle \sf \frac{dy}{dx} = 3.\sin\left(1-2x\right)^{3-1}.\left(\sin\left(1-2x\right)\right)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = 3.\sin\left(1-2x\right)^{2}.\left(\cos\left(1-2x\right)\right).\left(1-2x\right)'$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = 3.\sin\left(1-2x\right)^{2}.\left(\cos\left(1-2x\right)\right).\left(-2\right)$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = -6.\sin\left(1-2x\right)^{2}.\left(\cos\left(1-2x\right)\right)$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} = -6.\sin^2\left(1-2x\right).\left(\cos\left(1-2x\right)\right)$

Hence we get required answer!