Basic calculus-2

Calculus Level 1

y = sin ⁡ 2 x 1 + cos ⁡ x \large y=\frac{\sin^2x}{1+\cos x}

Find d y d x \dfrac {dy}{dx} .

cos ⁡ x \cos x 1 − cos ⁡ x 1-\cos x sin ⁡ x \sin x sec ⁡ x \sec x sin ⁡ x − sec ⁡ x \sin x-\sec x tan ⁡ x \tan x

Mohammad Khaza by Mohammad Khaza , 20, Bangladesh

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2 solutions

Mohammad Khaza
Oct 4, 2017

shaping that, y = s i n 2 x 1 + c o s x y=\frac{sin^2x}{1+cosx}

or, y = 1 − c o s 2 x 1 + c o s x y=\frac{1-cos^2x}{1+cosx} = 1 − c o s x 1-cosx

now, d y d x = 0 + s i n x = s i n x \frac{dy}{dx}=0+sinx=sinx

3 Helpful 0 Interesting 0 Brilliant 0 Confused
Munem Shahriar
Jan 16, 2018

y = sin ⁡ 2 x 1 + cos ⁡ x y = 1 − cos ⁡ x 2 1 + cos ⁡ x 2 [ sin ⁡ 2 x = 1 − cos ⁡ 2 x ] y = − cos ⁡ x + 1 \begin{aligned} y & = \dfrac{\sin^2 x}{1+\cos x} \\ y & = \dfrac{1-\cos x^2}{1+\cos x^2} ~~~~~~~~[\sin^2 x = 1- \cos^2 x] \\ y & = -\cos x +1 \\ \end{aligned}

Now,

d d x ( 1 − cos ⁡ x ) \dfrac{d}{dx} (1-\cos x)

= 0 − ( − sin ⁡ x ) = 0 - (-\sin x)

= 0 + sin ⁡ x = 0+ \sin x

= sin ⁡ x = \sin x

2 Helpful 0 Interesting 0 Brilliant 0 Confused

thanks for posting a solution.

Mohammad Khaza - 3 years, 4 months ago

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