Question

differentiate: y=sec (theta) tan (theta)

Answer 1

do you know the product rule? u'v+uv'

Answer 2

i got \[y'=(\sec \theta)(\tan \theta)'+(\tan \theta)(\sec \theta)'\]

Answer 3

yeah, ideally you'd keep it consistent. so:
\(\dfrac{d}{dx}(f(x)g(x))=f'(x)g(x)+f(x)g'(x)
\\y'=[\dfrac{d}{dx}(sec (\theta))* tan (\theta)]+[sec (\theta))*\dfrac{d}{dx} tan (\theta)]
\)
what's the derivative of sec x?
tan x?

Answer 4

\[(secx)'=secx tanx\]
\[(tanx)'=\sec ^{2}x\]

Answer 5

replace, multiply, factor if possible

Answer 6

i did replace them now im stuck factoring
\[(\sec \theta)(\sec ^{2}\theta)\]

Answer 7

\[\sec^2(\theta) = 1 + \tan^2(\theta)\]

Answer 8

sorry, I was away

Answer 9

y=sec (theta) tan (theta)

y'=sec (theta)' tan (theta) + sec (theta) tan (theta)'
(gonna start replacing theta with x, cause I'm lazy)
=secxtanx(tanx) + secx*sec^2x
=secxtan^2x+ secx * sec^2x
=secx(tan^2x+sec^2x)

Answer 10

you can rewrite it further from here, I guess
\(1+tan^2x=sec^2x\)

\(secx(tan^2x+sec^2x)=secx(tan^2x+1+tan^2x)=sec(x)(1+2tan^2(x))\) although idk why

Answer 11

why you'd want to*9