Question 287939
(a) The derivative of tanu = sec^2 u du
in our case u = 2x, so du = 2 dx
our answer is
2*sec^2(2x)
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(b) this is a power rule:
the derivative of (ax+b)^n = n(ax+b)^(n-1)* d(ax+b)dx
in our case we get
6(5x+3)^5*5 = 30(5x+3)^5
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