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How would I differentiate cos(2x)/x^1/2

So for this question you can use either the product rule or the quotient rule and I'll run through them both.

First the quotient rule:

The quotient rule says that if you have h(x)=f(x)/g(x)

Then h'(x) = (f'(x)g(x)-f(x)g'(x))/(g(x))^2

So using f(x)=cos(2x) and g(x)=x^1/2

then f'(x)=-2sin(2x) and g'(x)=1/2x^-1/2

Plugging this into our formula gives us

h(x) = (-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Always remember to simplify afterwards which gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

Second the product rule:

What the product rule says is that if

h(x) = f(x)g(x)

then h'(x) = f(x)g'(x) + f'(x)g(x)

So if we say that h(x) = cos(2x)/x^1/2

Then we can say that f(x) = cos(2x) and g(x) = x^-1/2

Using the product rule we have:

f(x) = cos(2x)       f'(x) = -2sin(2x)

g(x) = x^-1/2      g'(x) = 1/2x^-3/2

So lastly we know that h(x) = f(x)g'(x) + f'(x)g(x)

So using what we've found out we can say that h(x) = (cos(2x))/(2x^3/2)-(2sin(2x))/x^1/2

Once again simplifying gives us

(-2x^1/2sin(2x)-1/2x^-1/2cos(2x))/x

George S. GCSE Further Mathematics  tutor, A Level Further Mathematic...

10 months ago

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