Find the slope of the graph of y = ln(f(2x)) at x = 1, where f(2) = 3 and f'(2) = -5

Two fairly difficult (for me!) calculus questions, would love it if someone could show me the process in solving it!

a) Find the slope of the graph of y = ln(f(2x)) at x = 1, where f(2) = 3 and f'(2) = -5

b) Find the value of the constant a such that f(x) = asin(2x) + xcosx has a critical point at x = pi

Let : y = ln [ ƒ(2x) ] .............................. (1)

and : ƒ(2) = 3, ƒ'(2) = -5 ...................... (2)
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From (1), diff ... w.r.t.x by Chain Rule,

dy/dx = { 1 / [ ƒ(2x) ] } · d/dx[ ƒ(2x) ]

. . . . .= { 1 / [ ƒ(2x) ] } · ƒ'(2x) · 2

. . . . .= [ 2· ƒ'(2x) ] / [ ƒ(2x) ] ................. (3)
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Hence, from (3), slope at x = 1 is

m = [ (dy/dx) at x = 1 ]

.... = [ 2· ƒ'(2) ] / [ ƒ(2) ]

.... = 2(-5) / (3)

.... = -10/3 ........................................… Ans.
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