\( \dfrac{d}{dx}[\log(4x^3 - 2x)] \) is equal to

Assessment: JAMB UTME - Mathematics - 2019 Subject: General Mathematics

Question 1 Report

\frac{12 x - 2}{4 x^{2}}

\frac{43 x^{2} - 2 x}{7 x}

\frac{4 x^{2} - 2}{7 x + 6}

\frac{12 x^{2} - 2}{4 x^{3} - 2 x}

Answer Details

To find d/dx[log(4x^3−2x)], we can apply the chain rule of differentiation. Let u = 4x^3−2x, then y = log(u). Applying the chain rule, we have: dy/dx = dy/du * du/dx To find du/dx, we need to differentiate u with respect to x, giving: du/dx = 12x^2 - 2 To find dy/du, we can use the formula for differentiating the natural logarithm, which gives: dy/du = 1/u Putting these together, we have: dy/dx = dy/du * du/dx = 1/u * (12x^2 - 2) = (12x^2 - 2)/(4x^3 - 2x) = 2(6x^2 - 1)/(2x(2x^2 - 1)) Therefore, the answer is (12x^2 - 2)/(4x^3 - 2x), which is equivalent to.

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Logarithms

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\( \dfrac{d}{dx}[\log(4x^3 - 2x)] \) is equal to

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