How do you differentiate $k (x) = - 3 cos x$ $k(x)=-3 cos x$ ?

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Answer 1 · 2016-10-26T21:21:22

You could start off by saying that:

y=k(x)

If this is the case:

$y = - 3 cos x$ $y=-3cosx$

$- \frac{1}{3} \cdot y = cos x$ $-1/3*y=cosx$

Now from here you can use implicit differentiation to get $\frac{d y}{d x}$ $(dy)/(dx)$ .

$- \frac{1}{3} \cdot \frac{d y}{d x} = - sin x$ $-1/3*(dy)/(dx)=-sinx$

Which means that:

$\frac{d y}{d x} = 3 sin x$ $(dy)/(dx)=3sinx$

And this means that:

$k'(x)=3sinx$

Realistically speaking, if you are studying differentiation, you've got to memorise:

If $f(x)=cosx$ , $f'(x)=-sinx$ .

This is very important.

How do you differentiate k(x)=-3 cos xk(x)=−3cosxk(x)=-3 cos x?