How do you take the derivative of # y=sec^2 x + tan^2 x#?

1 Answer
Stefan V.
Aug 25, 2015

#y^' = 4sec^2(x)tan(x)#

Explanation:

Assuming that you're aware of the fact that

#color(blue)(d/dx(tanx) = sec^2x)" "# and #" "color(blue)(d/dx(secx) = secx * tanx)#

you can differentiate this function by using the chain rule twice, once for #u^2#, with #u = secx# and once for #t^2#, with #t = tanx#.

This will get you

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

#y^' = d/(du)(u^2) * d/dx(u) + d/(dt)t^2 * d/dx(t)#

#y^' = 2u * d/dx(secx) + 2t * d/dx(tanx)#

#y^' = 2sec(x) * sec(x) * tan(x) + 2tan(x) * sec^2(x)#

This is equal to

#y^' = 2sec^2(x)tan(x) + 2sec^2(x)tan(x)#

#y^' = color(green)(4sec^2(x)tan(x))#

Related questions
See all questions in Derivatives of y=sec(x), y=cot(x), y= csc(x)
Impact of this question
1233 views around the world
You can reuse this answer
Creative Commons License