# How do you differentiate f(x)=((18sin(x))/(4+x^(-2))) using the quotient rule?

Apr 1, 2016

Quotient rule states that if $f \left(x\right) = g \frac{x}{h \left(x\right)}$, then $f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)}$

$g ' \left(x\right) = \frac{d}{\mathrm{dx}} 18 \sin \left(x\right) = 18 \cos \left(x\right)$ from trig identities and derivatives. You should have those memorised.

$h ' \left(x\right) = \frac{d}{\mathrm{dx}} \left(4 + {x}^{-} 2\right) = 0 - 2 {x}^{-} 3$
$= - 2 {x}^{-} 3$ or $= - \frac{2}{x} ^ 3$

I got this above value from simple differentiation and laws of indices.

${h}^{2} \left(x\right) = {\left(4 + {x}^{-} 2\right)}^{2}$
$= 16 + 8 {x}^{-} 2 + {x}^{4}$

From expanding brackets and laws of indices.

Make sure you understand how I arrived at all of these values first before going on. I've tried to make it straightforward, but it's not the clearest thing in the world.

Putting all of these values together in the equation at the top, you should get:

$f ' \left(x\right) = \frac{g ' \left(x\right) h \left(x\right) - g \left(x\right) h ' \left(x\right)}{{h}^{2} \left(x\right)} = \frac{18 \cos \left(x\right) \cdot \left(4 + {x}^{-} 2\right) + 2 {x}^{-} 3 \cdot 18 \sin \left(x\right)}{16 + 8 {x}^{-} 2 + {x}^{4}}$

From there and using some algebra and expanding brackets you can simplify the answer and change it around how you like.

Hope that makes sense.