Question

How would you differentiate this...

Answer 1

\[y =-\cos x \ln (secx + tanx)\]

Answer 2

Product rule and chain rule

Answer 3

How?

Answer 4

Start off by using the product rule and then you'll see where the chain rule comes in

Answer 5

Ok thanks!

Answer 6

product rule
\[(fg)' = f'*g + f*g'\]

Answer 7

I know what the product rule means but thanks anyways.

Answer 8

While you're doing the P.Rule you'll see you will need to do the C.Rule too.

Answer 9

Have a look if u want :)

Answer 10

so it would basically be: \[y=\sin x \ln (\sec x+\tan x)-\cos x(\sec x \tan x+\sec^2x)/(\sec x+\tan x)\] ?

Answer 11

@oldrin.bataku @dan815

Answer 12

$$y=-(\cos x)\log(\sec x+\tan x)\\y'=(\sin x)\log(\sec x+\tan x)-(\cos x)\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}$$

Answer 13

forgot the prime there... thanks!

Answer 14

Finally it would be
y'=(sinx)log(secx+tanx)-1

Answer 15

yup, got that! thanks!

Answer 16

on a side note: how would you differentiate \[y = x^m\] twice?

Answer 17

Indeed:$$(\cos x)\frac{\sec x\tan x+\sec^2 x}{\sec x+\tan x}=\frac{\sec x+\tan x}{\sec x+\tan x}=1$$

Answer 18

is it \[y = m x^{m-1}\]

Answer 19

Differentiate it twice:$$y=x^m\\y'=mx^{m-1}\\y''=m(m-1)x^{m-2}$$

Answer 20

just making sure... thanks!

Answer 21

when u plug that into an equation like \[x^2y \prime \prime - 7xy \prime +15y=0\]...how would you start to simplify it?

Answer 22

@oldrin.bataku