How would I take the derivative of #x^2ln(sin4x)#?

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Answer 1 · 2015-06-05T22:48:14

#y(x) = x^2ln(sin4x)#

You would use the product rule and several chain rules here.

Product Rule
#d/(dx)[f(x)g(x)] = f(x)(dg(x))/(dx) + g(x)(df(x))/(dx)#

# = f(x)g'(x) + g(x)f'(x)#

Chain Rule
#d/(dx)[g(h(x))] = (dg(h(x)))/(dx)*(dh(x))/(dx) = g'(h(x))*h'(x)#

#(dy)/(dx) = x^2*1/(sin4x)*overbrace(cos4x*4)^("Chain Rule") + ln(sin4x)*2x#

# = 4x^2cot4x + 2xln(sin4x)#