Question

$\sf\blue{\colorbox{pink}{DERIVATIVES}}$

Differentiate the following w.r.t.x. :-
$\sf{sin ( 1 - 2x) ^2 }$

▪ wanted answers with step - by - step - explanation ✔
▪ no irrelevant answers / copied answers ❌
▪ those who can't understand they no need to mention in the answer slot otherwise account will be reported ⚠

Answer 1

y= $\sf{sin ( 1 - 2x) ^2 }$

$\bf\frac{dy}{dx}$ = $\bf\frac{d [sin(1-2x)²]}{dx}$

By chain rule,

$\bf\frac{dy}{dx}$

= $\bf\frac{d [sin(1-2x)²]}{d(1-2x)²}$• $\bf\frac{d(1-2x)²}{d(1-2x)}$•$\bf\frac{d(1-2x)}{dx}$

= cos(1-2x)² •2(1-2x) •(-2)

= -4(1-2x)•cos(1-2x)²

Or, 4(2x-1)cos(1-2x)² ans.

Answer 2

Answer:

y= \sf{sin ( 1 - 2x) ^2 }sin(1−2x)

2

\bf\frac{dy}{dx}

dx

dy

= \bf\frac{d [sin(1-2x)²]}{dx}

dx

d[sin(1−2x)²]

By chain rule,

\bf\frac{dy}{dx}

dx

dy

= \bf\frac{d [sin(1-2x)²]}{d(1-2x)²}

d(1−2x)²

d[sin(1−2x)²]

• \bf\frac{d(1-2x)²}{d(1-2x)}

d(1−2x)

d(1−2x)²

•\bf\frac{d(1-2x)}{dx}

dx

d(1−2x)

= cos(1-2x)² •2(1-2x) •(-2)

= -4(1-2x)•cos(1-2x)²

Or, 4(2x-1)cos(1-2x)² ans.