How do you differentiate #y = 3x cos^2 (x)#?

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How do you differentiate #y = 3x cos^2 (x)#?

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Answer 1 · 2018-02-01T22:58:07

#3cosx#[#cosx-2xsinx#]

Explanation:

#d#[#uv#]= #vdu+udv# this is the product rule, where v and u are both functions of x.

Let u =#3x# and #v=cos^2x#

so we have ......#cos^2x#.[#3]#+#3x#[#-2sinxcosx#]

=#3cos^2x#-#6xsinxcosx#.....=#3cosx[cosx-2xsinx#].

#cos^2x#=#[cosx]^2# and so need to use the chain rule to differentiate this, =#2cosx#, times the derivative of #cosx# which is -#sinx#, so# d/dxcos^2x##=-2sinxcosx#. hope this helped.

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How do you differentiate #y = 3x cos^2 (x)#?

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