anonymous
  • anonymous
How would you differentiate this...
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[y =-\cos x \ln (secx + tanx)\]
anonymous
  • anonymous
Product rule and chain rule
anonymous
  • anonymous
How?

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anonymous
  • anonymous
Start off by using the product rule and then you'll see where the chain rule comes in
anonymous
  • anonymous
Ok thanks!
dumbcow
  • dumbcow
product rule \[(fg)' = f'*g + f*g'\]
anonymous
  • anonymous
I know what the product rule means but thanks anyways.
anonymous
  • anonymous
While you're doing the P.Rule you'll see you will need to do the C.Rule too.
anonymous
  • anonymous
Have a look if u want :)
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anonymous
  • anonymous
so it would basically be: \[y=\sin x \ln (\sec x+\tan x)-\cos x(\sec x \tan x+\sec^2x)/(\sec x+\tan x)\] ?
anonymous
  • anonymous
@oldrin.bataku @dan815
anonymous
  • anonymous
$$y=-(\cos x)\log(\sec x+\tan x)\\y'=(\sin x)\log(\sec x+\tan x)-(\cos x)\frac{\sec x\tan x+\sec^2x}{\sec x+\tan x}$$
anonymous
  • anonymous
forgot the prime there... thanks!
anonymous
  • anonymous
Finally it would be y'=(sinx)log(secx+tanx)-1
anonymous
  • anonymous
yup, got that! thanks!
anonymous
  • anonymous
on a side note: how would you differentiate \[y = x^m\] twice?
anonymous
  • anonymous
Indeed:$$(\cos x)\frac{\sec x\tan x+\sec^2 x}{\sec x+\tan x}=\frac{\sec x+\tan x}{\sec x+\tan x}=1$$
anonymous
  • anonymous
is it \[y = m x^{m-1}\]
anonymous
  • anonymous
Differentiate it twice:$$y=x^m\\y'=mx^{m-1}\\y''=m(m-1)x^{m-2}$$
anonymous
  • anonymous
just making sure... thanks!
anonymous
  • anonymous
when u plug that into an equation like \[x^2y \prime \prime - 7xy \prime +15y=0\]...how would you start to simplify it?
anonymous
  • anonymous
@oldrin.bataku

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