Question

please someone explain to me on how to find the derivative of y= 4sec^2x

Answer 1

@ICECREAM_HERO do you know basic differentiation?

Answer 2

yup I just don't know how my teacher used the chain rule on this to get the answer o.O

Answer 3

ok, it'd work like this
\[\frac{d}{dx} (f(g(x))=f'(x)\times g'(x)\]
I'll illustrate with an example
\[\frac{d}{dx} (\sin ^2x)=2\times \sin x\times \frac{d}{dx} ( \sin x)\]
first I found derivative of the square term, works the same as x^2=> 2x
then I multiplied it with the derivative of sin x
did you understand this?

Answer 4

yup

Answer 5

could you apply this method to your question?

Answer 6

I'll try, I know how to use the chain rule in other problems but this one was just weird. The end answer is suppose to be \[8\sec^2xtanx\] but I keep getting something different

Answer 7

You'd get this only
\[\frac{d}{dx}4\times \sec^2 x\]
\[4\frac{d}{dx} \sec^2 x\]
Let sec x= f(x)
then we have
\[4\frac d {dx} (f(x))^2\]
\[4 \times (2 f(x))\frac d {dx} f(x)\]
now put f(x) 's value
\[4\times (2\times \sec x)\frac {d}{dx} (\sec x)\]
do you get this?

Answer 8

sort of so on the last equation at the last part is that just taking the derivative of sec x? and how did you get \[(d/dx) 4 * \sec^2x\] sorry this is why I need help

Answer 9

I chose f(x)= sec x
so we needed to find
\[\frac {d}{dx } (f(x))^2\]
The first step is just like finding derivative of x^2=> 2x
it's multiplied by the derivative of f(x) i.e f'(x)
\[2 f(x) f'(x)\]

Answer 10

OHHHH otay I see o.o

Answer 11

Do you get it through?

Answer 12

better than before it's just a different teaching style so how would you get the tanx in the solution

Answer 13

derivative of sec x is
\[\frac{d}{dx} (\sec x)=\sec x\times \tan x\]

Answer 14

HAHA OHH never mind sorry that was on my formula sheet XD should've known that and thank you so much!! I actually understand this better

Answer 15

Welcome :D