2 derivative questions ?

Question

Differentiate twice the following expression:

1/x^59

Differentiate once the following expression:

y=sin(2x)cos(3x)

Thank you

Answer 1

1) Convert to negative exponents first. It's easier. Then use the Power Rule.

1/x^59 = x^-59

d/dx (x^-59) = -59x^-60

Then once more:

d/dx (-59x^-60) = 3540x^-61 = 3540/x^61

2) You need to use a combination of the Product Rule and Chain Rule here. Also know that sin'(x) = cos(x) and cos'(x) = -sin(x). So:

y=sin(2x)cos(3x)

y'=sin'(2x)cos(3x) + cos'(3x)sin(2x)

y'=2cos(2x)cos(3x) - 3sin(3x)sin(2x)

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-IMP ;) :)

Answer 2

   y = 1/x^59

   y =  x^ -59                n.b. : (x^n)' = nx^(n-1)

1) y' = (x^ -59)' = -59x^ -60

2) y'' = (-59x^ -60)' = (-59)(-60)x^ -61 = 3540x^ -61

y = sin(2x)cos(3x)

y' = (sin(2x))' (cos(3x)) + (cos(3x))' (sin(2x))

   =  (cos(2x)) (2x)' (cos(3x)) + (-sin(3x)) (3x)' (sin(2x))

   =  2cos(2x) (cos(3x)) -3sin(3x) (sin(2x))