Asked by Sam

Can someone help me find the derivatives of these questions?

For these can you only need to tell me the answer so i can see if i have the same thing.
y=cos(-4x)

y = sinx^2

y = xcosx


I need help with these 6 please.

y = xcos(1/x)

y = (1+sinx)/(1-sin^(2)x)

y = sin^2x/cosx

y = sqrt(sin(sqrtx))

y = 3sqrt(xcosx)

y = 1/(sin(x-sinx)

Answers

Answered by Reiny
y=cos(-4x)
y' = -4(-sin(-4x)
= 4sin(-4x)

y = sinx^2 or (sinx)^2

y' = 2sinxcosx or sin 2x


y = xcosx
y' = x(-sinx) + (1)cosx
= cosx - xsinx
Answered by Reiny
for y = xcos(1/x)
I would use the product rule

y' = x(-sin(1/x))(-1/x^2) + (1)cos(1/x)



for y = (1+sinx)/(1-sin^(2)x) I would reduce it to

y = (1+sinx)/((1+sinx)(1-sinx))
= 1/(1-sinx)
= (1-sinx)^-1

now y' = -1(1-sinx)<sup>-2?sup>(-cosx)

why don't you show me some of the steps you have sofar for the others?
Answered by Reiny
last line should have come out as ...

y = -1(1-sinx)<sup>-2</sup>(-cosx)
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