How do you find the derivative of tanx using the product rule

best correct answer gets 10 points!

tanx = sinx/cosx
tanx = sinx * (cosx)⁻¹

Note: Never write 1/cosx as cos⁻¹x. This is the inverse cos function and is not the same as 1/cosx or (cosx)⁻¹

d/dx (tanx) = d/dx (sinx) * (cosx)⁻¹ + sinx * d/dx ((cosx)⁻¹)
. . . . . . . . . = cosx * (cosx)⁻¹ + sinx * -1 (cosx)⁻² (-sinx)
. . . . . . . . . = 1 + sin²x / cos²x
. . . . . . . . . = (cos²x + sin²x) / cos²x
. . . . . . . . . = 1/cos²x
. . . . . . . . . = sec²x


Or using quotient rule:

tanx = sinx/cosx

d/dx (tanx) = (d/dx (sinx) * cosx - sinx * d/dx (cosx)) / cos²x
. . . . . . . . . = (cosx * cosx - sinx * (-sinx)) / cos²x
. . . . . . . . . = (cos²x + sin²x) / cos²x
. . . . . . . . . = 1/cos²x
. . . . . . . . . = sec²x


Mαthmφm

sin(x) * 1/cos(x) = tan(x)
sin(x) * sec(x) = tan(x)

(uv)' = u'v + uv'

[sin(x)]' * sec(x) + sin(x) * [sec(x)]'
cos(x) * sec(x) + sin(x) * sec(x)tan(x)
1 + [sin(x) * sec(x)tan(x)]
1 + [sin(x) * 1/cos(x) * sin(x)/cos(x)]
1 + {[sin^2(x)]/[cos^2(x)]}
1 + tan^2(x) = sec^2(x)

That last step is a trig identity.