# How do you use the quotient rule to find the derivative of y=(1+cos(x))/(1+sin(x)) ?

##### 1 Answer
Sep 9, 2014

The quotient rule states that given functions u and v such that $y = \frac{u \left(x\right)}{v \left(x\right)} , \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{u ' \left(x\right) v \left(x\right) - u \left(x\right) v ' \left(x\right)}{{v}^{2} \left(x\right)}$ By assigning u and v equal to the numerator and denominator, respectively, in the example function, we arrive at $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- \sin x\right) \left(1 + \sin x\right) - \left(1 + \cos x\right) \left(\cos x\right)}{1 + \sin x} ^ 2$

When using the Quotient Rule, it is important to designate your functions $u \left(x\right)$ and $v \left(x\right)$ in such a way as to make things simple while still being accurate. In the above case, declaring $\cos \left(x\right)$ as $u \left(x\right)$ would not allow us to use the quotient rule efficiently, as our numerator would then be $1 + u \left(x\right)$.

From the identities of trigonometric function derivatives, we know that the derivative of $\sin \left(x\right)$ is $\cos \left(x\right)$, and the derivative of $\cos \left(x\right)$ is $- \sin \left(x\right)$. This could be proven using Euler's Formula, but for our purposes we shall accept these without proof. The derivative of any constant (such as 1 in our example) is 0, and the derivative of a sum is equal to the sum of the derivatives. Therefore:

$u \left(x\right) = 1 + \cos \left(x\right) , u ' \left(x\right) = - \sin \left(x\right) , v \left(x\right) = 1 + \sin \left(x\right) , v ' \left(x\right) = \cos \left(x\right)$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(- \sin x\right) \left(1 + \sin x\right) - \left(1 + \cos x\right) \left(\cos x\right)}{1 + \sin x} ^ 2$

Attempting to use trig identities to simplify...

$= \frac{- \sin \left(x\right) - {\sin}^{2} \left(x\right) - \cos \left(x\right) - {\cos}^{2} \left(x\right)}{1 + \sin x} ^ 2 = \frac{- 1 \left({\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)\right) - 1 \left(\sin x + \cos x\right)}{1 + \sin x} ^ 2 = - \frac{1 + \sin \left(x\right) + \cos \left(x\right)}{1 + \sin \left(x\right)} ^ 2$

However, the initial answer should be sufficient.