recall that if y = secx
dy/dx = secx tanx
y= [(sec^7 x)/7] - [(sec^5 x)/5]
dy/dx = (1/7)(7)(sec^6 x)(secx)tanx - (1/5)(5)(sec^4 x)secx tanx
= sec^7 x tanx - sec^5 x tanx
= sec^5 x tanx(sec^2 x - 1) but sec^2 x - 1 = tan^2x
= (sec^5 x)(tan^3 x)
nice question
Hi. I need help with finding the derivative of the function:
y= [(sec^7 x)/7] - [(sec^5 x)/5]
Thanks!
5 answers
Thank you Reiny!
I'm a little confused though.. Where did you get the (1/7) and (1/5) in the second line?
the original fractions had 7 and 5 in the denominator
which is the same as (1/7) times ....
e.g
9/4 = (1/4)(9)
which is the same as (1/7) times ....
e.g
9/4 = (1/4)(9)
Ohh I see. Thank you!