Question

how do i find the derivative of:

Y=x^3 tan^2 x

Answer 1

you have a function times a function so you have to use the product rule

Answer 2

so we have y'=[x^3]'(tanx)^2+x^3[(tanx)^2]'

Answer 3

(x^3)'=3x^2 using power rule

Answer 4

we have [(tanx)^2]'=2(tanx)*[secx]^2 using chain rule

Answer 5

thank you so much

Answer 6

np

Answer 7

\[3x^2(\tan^2x)+x^3(1/\cos^2x)\]

Answer 8

thank you

Answer 9

anytime :)

Answer 10

That's not right mmonish91

Answer 11

we know the equation is uv' + u'v, so take the following:
- u = x^3
- u' = 3x^2
- v = tan^2x
- v' = 2 tanx sec^2x

so:

y' = uv' + u'v
= (x^3)(2tanxsec^2x) + (3x^2)(tan^2x)

Answer 12

^_^ simply as that.

Answer 13

simple*

Answer 14

@myinninaya please do explain

Answer 15

mmonish dear, you started correctly, but didn't get d/dx tan^2x right.

the derivative of tanx = sec^2x , but in this case the derivative of:

(tanx)^ 2 = 2.(tanx)(sec^2x) ; you multiply the power first then subtract it, then you derive tanx. (power rule) ^_^

a^ u = u(a)(u') <-- rule :)

Answer 16

i see...i left out the 2 for tan^2
mah bad

Answer 17

a^ u = u(a)^u-1 (u') , sorry lol

Answer 18

no problem, we all make mistakes ^_^

Answer 19

i don't really work with sec so my professor said we could leave it in sin or cos format

Answer 20

but you must know the derivative of tanx = sec^2x

while the derivative of secx = secxtanx :)

Answer 21

which is same as 1/cos^2x

Answer 22

but I see what you mean :), u left it in cos, it's alright

Answer 23

yes, true :)

Answer 24

thank you guys!! ^.^

Answer 25

you are most welcome sindy ^_^ glad you understood it

Answer 26

yes i did understand :D ..