Question

differential of sin^2[ln(sec)]​

Answer 1

Given :

y = sin²θ.ln(secθ)

To Find :

dy/dθ

Solution :

$\bf y=sin^2\theta.ln(sec\theta)\\\\\implies \bf \dfrac{dy}{d \theta}=ln(sec\theta).\dfrac{d}{d\theta}(sin^2\theta)+sin^2\theta\dfrac{d}{d\theta}(ln(sec\theta))\\\\\implies \bf \dfrac{dy}{d\theta}=ln(sec\theta).2sin\theta.cos\theta+sin^2\theta.\dfrac{1}{sec\theta}.sec\theta.tan\theta\\\\\implies \bf \dfrac{dy}{d\theta}=sin2\theta.ln(sec\theta)+sin^2\theta.tan\theta$

Used Formulas :

$\bullet \;\;\bf \dfrac{d}{dx}(uv)=v\dfrac{du}{dx}+u\dfrac{dv}{dx}\\\\\bullet \;\; \bf \dfrac{d}{dx}(lnx)=\dfrac{1}{x}\\\\\bullet \;\; sin2x=2sinx.cosx\\\\\bullet \;\; \dfrac{d}{dx}(secx)=secx.tanx$

Answer 2

Answer:

it can be solved by breaking it into small parts.

Step-by-step explanation:

let M = sin^2[ln(sec)], now Y = ln(secx),

now M can be writtten as sin^2(Y)

we know that,

sin^2(x) = 0.5*(1 - cos2x)

this implies, M = 0.5*(1 - cos2Y)

now differentiate M,

d(M)/dx = 0.5*[0 - (-sin2Y).d(2Y)/dx]

d(M)/dx = sin2Y.d(Y)/dx

now d(Y)/dx = d(ln(secx))/dx

                    = (1/secx).d(secx)/dx

                    = (1/secx).(secx.tanx)

     d(Y)/dx   = tanx

therefore,

d(M)/dx = sin2{ln(secx)}.tanx