How do you differentiate $f (x) = (4 - x^{2}) \cdot ln x$ $f(x)= (4-x^2) *ln x$ using the product rule?

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Answer 1 · 2016-01-02T03:15:53

$\frac{(4 - x^{2}) - 2 x^{2} \cdot ln x}{x}$ $((4-x^2)-2x^2 * lnx)/x$

Product rule: $h = f \cdot g$ $h = f*g$

$h'= fg'+gf'$

Note: $f(x) = ln x$

$f'(x) = 1/x$

Given $f(x) = (4-x^2)*lnx$

$f'(x) = (4-x^2) d/dx(lnx) + lnx *d/dx(4-x^2)$

$= (4-x^2) (1/x) + -2x(lnx)$

$= (4-x^2)/x - (2x) (ln x)$

= $((4-x^2)-2x^2 * lnx)/x$

How do you differentiate f(x)= (4-x^2) *ln xf(x)=(4−x2)⋅lnxf(x)= (4-x^2) *ln x using the product rule?