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# Integration of Trigonometric fucntions watch

1. Given that y = (secx+tanx)^2

show that dy/dx = 2 y secx

i have managed to do this bit

find d^2y/dx^2 , in terms of y, secx and tanx - is this bit correct?

d^2y/dx^2 = 4 (secx + tanx) (secxtanx + tanx) secxtanx

show that cosx(d^2y/dx^2) - 2(dy/dx) = 2 y tanx

so....

4 cosxsecxtanx (tanxsecx + secx)(secx +tanx) - 2(secx +tanx)^2 secx= LHS

where do i go from here???
2. (Original post by Custardcream000)
find d^2y/dx^2 , in temrs of y =secx and tanx - is this bit correct?
Please post the question as it was written.
3. (Original post by Custardcream000)
Given that y = (secx+tanx)^2

show that dy/dx = 2 y secx

i have managed to do this bit

find d^2y/dx^2 , in temrs of y =secx and tanx - is this bit correct?

d^2y/dx^2 = 4 (secx + tanx) (secxtanx + tanx) secxtanx

show that cosx(d^2y/dx^2) - 2(dy/dx) = 2 y tanx

so....

4 cosxsecxtanx (tanxsecx + secx)(secx +tanx) - 2(secx +tanx)^2 secx= LHS

where do i go from here???
Looks absolutely hideous - keep it in terms of y as required by the question, please. Your second derivative is wrong.

So we know

Therefore by the chain and product rules we get and you can sub for dy/dx and keep it in terms of y.
4. so for iii)

LHS = cosx(2ysecxtanx + 2(dy/dx)secx) -2(2ysecx)
= 2ysecx (tanxcosx + secx - 2)

if you cnace whats in the bracket = sinx -2

so you would get 2ytanx put you would be left with -2

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