How do you differentiate $f (x) = \frac{x^{2} - 4 x}{x cot x + 1}$ $f(x) = (x^2-4x)/(xcotx+1)$ using the quotient rule?

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How do you differentiate $f (x) = \frac{x^{2} - 4 x}{x cot x + 1}$ $f(x) = (x^2-4x)/(xcotx+1)$ using the quotient rule?

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Answer 1 · 2017-07-21T18:45:07

$\frac{d f}{d x} = \frac{(2 x - 4) (x cot x + 1) + (\frac{x}{{sin}^{2} θ} - cot x) (x^{2} - 4 x)}{{(x cot x + 1)}^{2}}$ $(df)/dx=((2x-4)(xcotx+1)+(x/sin^2theta-cotx)(x^2-4x))/(xcotx+1)^2$

Explanation:

$\frac{d f}{d x} = \frac{\frac{d (x^{2} - 4 x)}{d x} (x cot x + 1) - \frac{d (x cot x + 1)}{d x} (x^{2} - 4 x)}{{(x cot x + 1)}^{2}} =$ $(df)/dx=((d(x^2-4x))/dx(xcotx+1)-(d(xcotx+1))/dx(x^2-4x))/(xcotx+1)^2=$

$\frac{(2 x - 4) (x cot x + 1) - (- \frac{x}{{sin}^{2} θ} + cot x) (x^{2} - 4 x)}{{(x cot x + 1)}^{2}} =$ $((2x-4)(xcotx+1)-(-x/sin^2theta+cotx)(x^2-4x))/(xcotx+1)^2=$

$\frac{(2 x - 4) (x cot x + 1) + (\frac{x}{{sin}^{2} θ} - cot x) (x^{2} - 4 x)}{{(x cot x + 1)}^{2}}$ $((2x-4)(xcotx+1)+(x/sin^2theta-cotx)(x^2-4x))/(xcotx+1)^2$

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How do you differentiate $f (x) = \frac{x^{2} - 4 x}{x cot x + 1}$ $f(x) = (x^2-4x)/(xcotx+1)$ using the quotient rule?

1 Answer

Explanation:

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How do you differentiate $f (x) = \frac{x^{2} - 4 x}{x cot x + 1}$ $f(x) = (x^2-4x)/(xcotx+1)$ using the quotient rule?