How do you differentiate $f (x) = \frac{sin x}{1 - cos x}$ $f(x)=sinx/(1-cosx)$ ?

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How do you differentiate $f (x) = \frac{sin x}{1 - cos x}$ $f(x)=sinx/(1-cosx)$ ?

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Answer 1 · 2016-10-05T10:03:23

$\frac{d f}{d x} = - \frac{1}{1 - cos x}$ $(df)/(dx)=-1/(1-cosx)$

Explanation:

We use the quotient formula here. It states if $f (x) = \frac{g (x)}{h (x)}$ $f(x)=(g(x))/(h(x))$

then $\frac{d f}{d x} = \frac{\frac{d g}{d x} \times h (x) - \frac{d h}{d x} \times g (x)}{{(h (x))}^{2}}$ $(df)/(dx)=((dg)/(dx)xxh(x)-(dh)/(dx)xxg(x))/(h(x))^2$

As $f (x) = \frac{sin x}{1 - cos x}$ $f(x)=sinx/(1-cosx)$

$\frac{d f}{d x} = \frac{cos x \times (1 - cos x) - sin x \times sin x}{{(1 - cos x)}^{2}}$ $(df)/(dx)=(cosx xx(1-cosx)-sinx xxsinx)/(1-cosx)^2$

= $\frac{cos x - {cos}^{2} x - {sin}^{2} x}{{(1 - cos x)}^{2}}$ $(cosx-cos^2x-sin^2x)/(1-cosx)^2$

= $\frac{cos x - ({cos}^{2} x + {sin}^{2} x)}{{(1 - cos x)}^{2}}$ $(cosx-(cos^2x+sin^2x))/(1-cosx)^2$

= $\frac{cos x - 1}{{(1 - cos x)}^{2}}$ $(cosx-1)/(1-cosx)^2$

= $- \frac{1 - cos x}{{(1 - cos x)}^{2}}$ $-(1-cosx)/(1-cosx)^2$

= $- \frac{1}{1 - cos x}$ $-1/(1-cosx)$

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How do you differentiate $f (x) = \frac{sin x}{1 - cos x}$ $f(x)=sinx/(1-cosx)$ ?

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Explanation:

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