How do you differentiate #4/cosx + 1/ tanx#?

Question

Elijah Abram · Accepted Answer

#d/dx(4secx + cotx)= 4secxtanx - csc^2x# #y=4/cosx+1/tanx# #dy/dx= d/dx(4secx + cotx) = 4secxtanx - csc^2x#

Evelyn Agard · Answer

undefined To differentiate the expression $ \frac{4}{\cos(x)} + \frac{1}{\tan(x)} $, you can use the quotient rule.

The quotient rule states that if you have a function $ \frac{u(x)}{v(x)} $, then its derivative is given by:

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

For $ u(x) = 4 $ and $ v(x) = \cos(x) $, we have $ u'(x) = 0 $ and $ v'(x) = -\sin(x) $. Similarly, for $ u(x) = 1 $ and $ v(x) = \tan(x) $, we have $ u'(x) = 0 $ and $ v'(x) = \sec^2(x) $.

Plugging these values into the quotient rule:

$$ \frac{d}{dx}\left(\frac{4}{\cos(x)} + \frac{1}{\tan(x)}\right) = \frac{0 \cdot \cos(x) - 4 \cdot (-\sin(x))}{(\cos(x))^2} + \frac{0 \cdot \tan(x) - 1 \cdot \sec^2(x)}{(\tan(x))^2} $$

$$ = \frac{4\sin(x)}{\cos^2(x)} - \frac{\sec^2(x)}{\tan^2(x)} $$

$$ = \frac{4\sin(x)}{\cos^2(x)} - \frac{1}{\sin^2(x)} $$

$$ = 4\sin(x)\csc^2(x) - \csc^2(x) $$

$$ = \csc^2(x)(4\sin(x) - 1) $$

So, the derivative of $ \frac{4}{\cos(x)} + \frac{1}{\tan(x)} $ with respect to $ x $ is $ \csc^2(x)(4\sin(x) - 1) $.