How do you find the derivative of # tanxsecx#?

Question

Charles Arocha · Accepted Answer

#d/(dx)(tan(x)sec(x)) = sec(x)(sec^2(x)+tan^2(x))#

Can sub in for the term in brackets to try and simplify but not really worth it imo

Ok, this is a relatively simple application of the product rule if we know the derivatives of tan(x) and sec(x) but we'll derive them to give it a bit of challenge. #color(blue)("Optional Derivation Interlude")# Remember that #tan(x) = (sin(x))/(cos(x))# #d/(dx)(tan(x)) = d/(dx)((sin(x))/(cos(x)))# Going to use the quotient rule here: #d/(dx)((f(x))/(g(x))) = (f'(x)g(x) - f(x)g'(x))/(g(x))^2# #d/(dx)(tan(x)) = (cos(x)cos(x) - sin(x)(-sin(x)))/(cos(x))^2 = (cos^2(x) + sin^2(x))/(cos^2(x))# Recall that #sin^2x + cos^2x = 1# so: #d/(dx)(tan(x)) = 1/(cos^2(x)) = sec^2(x)# #d/(dx)(sec(x)) = d/(dx)(1/(cos(x)))# Quotient rule time again: #(0*cos(x) - 1*(-sin(x)))/(cos^2(x)) = (sin(x))/(cos^2(x)) = (sin(x))/(cos(x))(1)/(cos(x))# #therefore d/(dx)(sec(x)) = tan(x)sec(x)# #color(blue)("Derivation End")# Product rule given by: #d/(dx)(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)# #d/(dx)(tan(x)sec(x)) = d/(dx)(tan(x))sec(x) + tan(x)d/(dx)(sec(x))# #=sec^2(x)sec(x) + tan(x)tan(x)sec(x)# #=sec^3(x) + tan^2(x)sec(x) = sec(x)(sec^2(x)+tan^2(x))#

Charles Anctil · Answer

undefined To find the derivative of $ \tan(x)\sec(x) $, you can use the product rule, which states that the derivative of the product of two functions is the derivative of the first function times the second function plus the first function times the derivative of the second function.

Using the product rule, the derivative of $ \tan(x)\sec(x) $ with respect to $ x $ is:

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

Simplify the expression:

$$ = \sec^3(x) + \tan(x)\sec^2(x)\tan(x) $$

$$ = \sec^3(x) + \tan^2(x)\sec^2(x) $$

So, the derivative of $ \tan(x)\sec(x) $ is $ \sec^3(x) + \tan^2(x)\sec^2(x) $.