What is the derivative of  y= 1/ (sinx+cosx)?

Jul 28, 2015

${y}^{'} = - \frac{\cos x - \sin x}{\sin x + \cos x} ^ 2$

Explanation:

Notice that your function is actually the quotient of two other functions, which means that you can use the quotient rule to determine its derivative.

More specifically, those two functions are

$f \left(x\right) = 1$ and $g \left(x\right) = \sin x + \cos x$

such that your function can be written as

$y = f \frac{x}{g} \left(x\right) = \frac{1}{\sin x + \cos x}$

Now, the quotient rule says that th derivative of a quotient of two functions is equal to

$\frac{d}{\mathrm{dx}} \left[f \frac{x}{g} \left(x\right)\right] = \frac{{f}^{'} \left(x\right) \cdot g \left(x\right) - f \left(x\right) \cdot {g}^{'} \left(x\right)}{g \left(x\right)} ^ 2$, where $g \left(x\right) \ne 0$

You also know that

$\frac{d}{\mathrm{dx}} \left(\sin x\right) = \cos x$

$\frac{d}{\mathrm{dx}} \left(\cos x\right) = - \sin x$

and that the derivative of a constant is equal to zero

$\frac{d}{\mathrm{dx}} \left(a\right) = 0$

which means that you have

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

${y}^{'} = \frac{0 \cdot \left(\sin x + \cos x\right) - \frac{d}{\mathrm{dx}} \left(\sin x\right) - \frac{d}{\mathrm{dx}} \left(\cos x\right)}{\sin x + \cos x} ^ 2$

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

${y}^{'} = \textcolor{g r e e n}{- \frac{\cos x - \sin x}{\sin x + \cos x} ^ 2}$

SIDE NOTE You could also use the reciprocal rule, which deals with the special case of $f \left(x\right) = 1$

$\frac{d}{\mathrm{dx}} \left(\frac{1}{g} \left(x\right)\right) = \frac{- g ' \left(x\right)}{g \left(x\right)} ^ 2$, where once again $g \left(x\right) \ne 0$.