How do you derive #y = (sinx)/[1 - sin^2(x))# using the quotient rule?

Question

Ezekiel Alexander · Accepted Answer

I found: #y'=(cos(x)[1+sin^2(x)])/([1-sin^2(x)]^2)=#

Given a function as #f(x)=(h(x))/g(x)# the Quotient Rule tells us that:

#y'=f'(x)=(h'(x)g(x)-h(x)g'(x))/([g(x)]^2)#

In your case you get:

#y'=f´(x)=(cos(x)(1-sin^2(x))-sin(x)(-2sin(x)cos(x)))/([1-sin^2(x)]^2)=# #=(cos(x)-cos(x)sin^2(x)+2sin^2(x)cos(x))/([1-sin^2(x)]^2)=# #=(cos(x)+cos(x)sin^2(x))/([1-sin^2(x)]^2)=# #=(cos(x)[1+sin^2(x)])/([1-sin^2(x)]^2)=#

Charles Anctil · Answer

undefined To derive $ y = \frac{\sin(x)}{1 - \sin^2(x)} $ using the quotient rule, we first identify $ u(x) = \sin(x) $ and $ v(x) = 1 - \sin^2(x) $. Then, we use the quotient rule formula:

$$ \frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{[v(x)]^2} $$

Now, we find the derivatives of $ u(x) $ and $ v(x) $:

$$ u'(x) = \cos(x) $$
$$ v'(x) = -2\sin(x)\cos(x) $$

Substituting these into the quotient rule formula:

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

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

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

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

So, the derivative of $ y = \frac{\sin(x)}{1 - \sin^2(x)} $ with respect to $ x $ using the quotient rule is:

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