How do you differentiate $f (x) = ln 2 x \cdot cot (5 - x)$ $f(x)=ln2x * cot(5-x)$ using the product rule?

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How do you differentiate $f (x) = ln 2 x \cdot cot (5 - x)$ $f(x)=ln2x * cot(5-x)$ using the product rule?

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Answer 1 · 2016-06-19T06:57:18

$\frac{d f}{d x} = \frac{cot (5 - x)}{x} + {csc}^{2} (5 - x) ln (2 x)$ $(df)/(dx)=cot(5-x)/x+csc^2(5-x)ln(2x)$

Explanation:

Product rule states if $f (x) = g (x) h (x)$ $f(x)=g(x)h(x)$

then $\frac{d f}{d x} = \frac{d g}{d x} \times h (x) + \frac{d h}{d x} \times g (x)$ $(df)/(dx)=(dg)/(dx)xxh(x)+(dh)/(dx)xxg(x)$

Hence as $f (x) = ln (2 x) cot (5 - x)$ $f(x)=ln(2x)cot(5-x)$

$\frac{d f}{d x} = \frac{1}{2 x} \times 2 \times cot (5 - x) - {csc}^{2} (5 - x) \times (- 1) \times ln (2 x)$ $(df)/(dx)=1/(2x)xx2xxcot(5-x)-csc^2(5-x)xx(-1)xxln(2x)$

or $\frac{d f}{d x} = \frac{cot (5 - x)}{x} + {csc}^{2} (5 - x) ln (2 x)$ $(df)/(dx)=cot(5-x)/x+csc^2(5-x)ln(2x)$

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How do you differentiate $f (x) = ln 2 x \cdot cot (5 - x)$ $f(x)=ln2x * cot(5-x)$ using the product rule?

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How do you differentiate f(x)=ln2x * cot(5-x)f(x)=ln2x⋅cot(5−x)f(x)=ln2x * cot(5-x) using the product rule?

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How do you differentiate $f (x) = ln 2 x \cdot cot (5 - x)$ $f(x)=ln2x * cot(5-x)$ using the product rule?