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# Maclaurin's theorem watch

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1. I was doing a question - Given that y = ln cos(2X), find d^4y/dX^4

I got dy/dx as -2sin2X/cos2X and d^2y/dX^2 as -4/cos^2(2X) (the markscheme put -4sec^2(2X) which I'm pretty sure is the same)

Then for d^3y/dX^3, I used the quotient rule to get d^3y/dX^3 =

cos(2X)(0) - -4(d/dX(cos^2(2X))) / (cos^2(2X))^2

for d/dX(cos^2(2X)), I put it equal to d/dX ((cos2X)^2) = -4sin(2X) x cos(2X)

Then I got d^3y/dX^3 = -16sin(2X) x (cos(2X)) / (cos(2X))^2, so I cancelled out the cos(2X) to get d^3y / dX^3 = -16sin(2X) / cos(2X) = -16tan(2X)

But the markscheme says it is -16tan(2X) x sec^2(2X)

Where have I gone wrong please???
2. (Original post by MathsALevel123)
I was doing a question - Given that y = ln cos(2X), find d^4y/dX^4

I got dy/dx as -2sin2X/cos2X and d^2y/dX^2 as -4/cos^2(2X) (the markscheme put -4sec^2(2X) which I'm pretty sure is the same)

Then for d^3y/dX^3, I used the quotient rule to get d^3y/dX^3 =

cos(2X)(0) - -4(d/dX(cos^2(2X))) / (cos^2(2X))^2

for d/dX(cos^2(2X)), I put it equal to d/dX ((cos2X)^2) = -4sin(2X) x cos(2X)

Then I got d^3y/dX^3 = -16sin(2X) x (cos(2X)) / (cos(2X))^2, so I cancelled out the cos(2X) to get d^3y / dX^3 = -16sin(2X) / cos(2X) = -16tan(2X)

But the markscheme says it is -16tan(2X) x sec^2(2X)

Where have I gone wrong please???
You just had a bit of a "mental block"

but you put

3. (Original post by Indeterminate)
You just had a bit of a "mental block"

but you put

Thanks!!!
4. (Original post by MathsALevel123)
Thanks!!!
My pleasure

Besides we've all made mistakes like that

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Updated: November 11, 2015
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