Question

ahoy!!

may I have your attention,please ?

read question and answer in full steps!

differentiate w.r.t x

=> cos. ( 3x - 5 )^2

answer should be -6 (3x +5) sin (3x + 5)^2

Answer 1

$\bf \: Hello \: Loco + Maniac$

$\mathfrak{Let \: me \: Clear \: a \: concept \: that's \: none \: other \: than \: } \\ \\ \bf \: chain \: rule :$
According to this we sum a fraction and after differentiation one by one from each side we multiply it.

Now coming to our Question :-

• Given :-

we need to differentiate with respect to x

cos ( 3x - 5 )²

$\boxed{ \bf \: Let's \: differentiate \: this :- }$

$cos. \frac{d( {3x - 5)}^{2} }{dx} + ({3x - 5})^{2}. \frac{d(cos)}{dx} \\ \\ \rightarrow \: -6( {3x - 5)} + ( sin {(3x - 5)}^{2} )$

Now according to our rule we get :-

$\rightarrow \: sin {(3x - 5)}^{2}.6(3x -5) \\ \\ \rightarrow \: \boxed{ \bf- 6(3x - 5)sin(3x -5) ^{2} }$

$\mathfrak {Good \: Luck } :)$

Answer 2

$\bf HELLO \: \: DEAR,$

$\bf let \: \: y = cos(3x - 5)^{2}$

$\boxed{\bf put \: \: \: (3x - 5) = t \: \:, y = cost^{2} }$


$\boxed{\bf put \: \: \: {t}^{2} = z \: \: , y = cosz}$

$\bf now \: (3x - 5) = t \: \: , \: \: t^{2} = z \: \: , \: \: y = cosz$

$\bf using \: \: chain \: \: rule \: \: \: dy/dx = dy/dz * dz/dt * dt/dx$


$\bf dy/dx= d(cosz)/dz * d(t^{2})/dt * d(3x - 5)/dx$

$\bf dy/dx = (-sinz) * (2t) * (3)$

$\bf dy/dx = -[sin(3x - 5)^{2} * 2(3x - 5) * (3)]$

$\bf dy/dx = -6(3x - 5) * sin(3x - 5)^{2}$



$\boxed{\underline { \bf \: I \: \: HOPE \: \: ITS \: \: HELP \: \: YOU \: \: DEAR, \: \: THANKS}}$