ChezCake14
Answered

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Find the derivative of x³ln(cos²x)

I just need the working and an explanation of how you got to your answer.

Sagot :

evgeniylevi

Try this option:

1) the rule 1: if f(x)=u(x)*v(x), then f'(x)=u'(x)*v(x)+u(x)*v'(x), where u(x)=x³, v(x)=ln(cos²x);

2) the rule 2: if f(x)=u(v(x)), then f'(x)=u'(v(x))*v'(x), where u(v(x))=ln(cos²x), v(x)=cos²x;

3) according to the rules above:

f'(x)=(x³)'*ln(cos²x)+x³*(ln(cos²x))';

[tex]f`(x)=3x^2ln(cos^2x)+x^3*\frac{1}{cos^2x}*(-2sinxcosx)=3x^2ln(cos^2x)-2x^3tanx.[/tex]

Answer: 3x²ln(cos²x)-2x³tanx.

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