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