A calculus derivative problem. Find y' of y= ln [sec5x + tan5x] "[ and ]" are standing for absolute value

Question

The correct answer is 5sec5x. How do you get this? Please show a lot of work.

Answer 1

The derivative of ln|f(x)| is 1/f(x) * d f(x) / dx; this will be

1/[sec5x + tan5x] * d(sec5x) / dx + d(tan5x) / dx

d(sec5x) / dx = d(1/cos5x) / dx = -1/cos^2(5x) * -5sin5x = 5tanx*secx

d(tan5x) / dx = 5sec^2(5x)

(5tanxsecx + 5sec^2(5x)) / (sec5x + tan5x)

Factor out 5sec5x to get 5sec5x * (tan5x + sec5x)/(sec5x + tan5x) = 5sec5x

Answer 2

attempt utilising the Chain rule the place f(x)=h(g(x)) and g(x)=sec5x +tan5x h(x)=ln(x) so ln(" ")(5(cos5x^2 + sin5x*sin5x+one million)/cos5x^2)+(one million/X)(" ") tried to variety it out yet in basic terms use the chin rule it could artwork.

Answer 3

Use the definition of the integral of sec(x).
=== int(sec(x)) = ln[sec(x) + tan(x)]

int(sec(x)) = ln [sec(x) + tan(x)]

thus you can say

int(sec(5x)) = ln [sec(5x) + tan(5x)] = y

thus y'

dx(int(sec(5x))) = sec(5x) = y'

y' = sec(5x)

If you didn't know this, doing it by hand can by kind of tricky.

You have to use a few tricks.

I'd write it out for you, but this site does it in a much more cleaner fashion.

http://math2.org/math/integrals/more/sec.htm