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How do you differentiate #f(x)=(-2x^2-5)/cos(2x^3)#?

Answer 1

Ezekiel Alexander

#f'(x) =(-4xcos(2x^3) - (12x^4+5)sin(2x^3))/cos^2(2x^3)#

Using the chain rule for the denominator:

#(A(B(x)))' = A'(B(x)) * B'(x)#

#(cos(2x^3))' = -sin(2x^3) * 6x^2#

Then, using the quotient rule:

#((A(x))/(B(x)))' = (A'(x)B(x) - A(x)B'(x))/(B^2(x))#

#(-4xcos(2x^3) - (-sin(2x^3)*6x^2)(-2x^2-5))/cos^2(2x^3)#

#=(-4xcos(2x^3) - (12x^4sin(2x^3) + 5sin(2x^3)))/(cos^2(2x^3)#

#=(-4xcos(2x^3) - (12x^4+5)sin(2x^3))/cos^2(2x^3)#

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Answer 2

William Abdalla

To differentiate (f(x)=\frac{-2x^2-5}{\cos(2x^3)}), you can use the quotient rule:

(f'(x) = \frac{(2\cos(2x^3)(-4x^2)-(-2x^2-5)(-6x^2\sin(2x^3)))}{\cos^2(2x^3)})

Simplify this expression for the derivative.

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Answer from HIX Tutor

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