How do you differentiate #f(t)=ln(e^(sin^2t))# using the chain rule?

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Answer 1 · 2016-04-24T15:19:03

Without rewriting

Use #d/dx lnu = 1/u (du)/dx# and #d/dx(e^u) = e^u (du)/dx# toi get

#f'(t) = 1/e^(sin^2t)[d/dt(e^(sin^2t))]#

# = 1/e^(sin^2t)[(e^(sin^2t)d/dt(sin^2t)]#

# = e^(sin^2t)/(e^(sin^2t))[2sintd/dx(sint))]#

# = 2sintcost = sin2t#

With rewriting

Use #ln(e^u) = u# to rewrite

#f(x) = sin^2x = (sinx)^2#

Now use the chain rule to get

#f'(x) = 2sint[d/dx(sint)] = 2sintcost = sin2t#