Derivative of sinx/(1+cosx) [sinx divided by 1+cosx]

The derivative of sinx/(1+cosx) [sinx divided by 1+cosx] is equal to 1/(1+cosx), and it is denoted by d/dx (sinx/1+cosx). That is,

$\dfrac{d}{dx} \big(\dfrac{\sin x}{1+\cos x} \big)=\dfrac{1}{1+\cos x}$.

Here we will learn how to differentiate sinx divided by 1+cosx.

Differentiate sinx/(1+cosx)

Question: If y= sinx/1+cosx, then find dy/dx.

Answer: If y=sinx/(1+cosx), then dy/dx = 1/(1+cosx).


Lets assume that

y = $\dfrac{\sin x}{1+\cos x}$

Cross-multiplying, we get that

(1+cosx) y = sinx

Now, differentiating both side with respect to x, we get that

(1+cosx) $\dfrac{dy}{dx}$ + y (0-sinx) = cosx, by the product rule of derivatives.

⇒ (1+cosx) $\dfrac{dy}{dx}$ – y sinx = cosx

⇒ (1+cosx) $\dfrac{dy}{dx}$ = cosx + ysinx

⇒ $\dfrac{dy}{dx} =\dfrac{\cos x + y\sin x}{1+\cos x}$

⇒ $\dfrac{dy}{dx} =\dfrac{\cos x + \dfrac{\sin x}{1+\cos x} \sin x}{1+\cos x}$, obtained by putting the value of y.

⇒ $\dfrac{dy}{dx} =\dfrac{\cos x(1+\cos x) + \sin x \cdot \sin x}{(1+\cos x)^2}$

⇒ $\dfrac{dy}{dx} =\dfrac{\cos x+\cos^2 x+ \sin^2 x}{(1+\cos x)^2}$

⇒ $\dfrac{dy}{dx} =\dfrac{1+\cos x}{(1+\cos x)^2}$ using the formula cos2x+sin2x=1.

⇒ $\dfrac{dy}{dx} =\dfrac{1}{1+\cos x}$.

So the derivative of sinx/(1+cosx), i.e., sinx divided by 1+cosx is equal to 1/(1+cosx).

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Q1: What is the derivative of sinx/1+cosx?

Answer: The derivative of sinx/1+cosx is equal to 1/(1+cosx).

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