How do you find the derivative of #f(x)= 5 sec(x) tan(x)# ?

Question

Cameron Alexander · Accepted Answer

By Product Rule, we can find 
#f'(x)=5secx(1+2tan^2x)#. Let us look at some details.
By pulling 5 out of the derivative,
#f'(x)=5[secxtanx]'#
by Product Rule,
#=5[secxtanx cdot tanx+secx cdot sec^2x]#
by factoring out #sec x#,
#=5secx(tan^2x+sec^2x)#
by #sec^2x=1+tan^2x#,
#=5secx(1+2tan^2x)#

Emily Ammerman · Answer

undefined To find the derivative of $ f(x) = 5\sec(x)\tan(x) $, you can use the product rule of differentiation. The product rule states that if $ u(x) $ and $ v(x) $ are both differentiable functions of $ x $, then the derivative of their product is given by $ (u v)' = u'v + uv' $. Applying this rule:

$$ f(x) = 5\sec(x)\tan(x) $$
$$ f'(x) = (5\sec(x))' \cdot \tan(x) + 5\sec(x) \cdot (\tan(x))' $$

Now, differentiate $ 5\sec(x) $ and $ \tan(x) $ with respect to $ x $:

$$ \frac{d}{dx}(5\sec(x)) = 5\sec(x)\tan(x) $$
$$ \frac{d}{dx}(\tan(x)) = \sec^2(x) $$

Substitute these derivatives back into the expression:

$$ f'(x) = (5\sec(x)\tan(x))' \cdot \tan(x) + 5\sec(x) \cdot (\sec^2(x)) $$
$$ f'(x) = (5\sec(x)\tan(x))' \cdot \tan(x) + 5\sec(x) \cdot \sec^2(x) $$

Simplify if necessary, and that's your derivative.