Calculus

Methods

The first-principles approach to finding a derivative is time consuming, and often messy or difficult.

As you continue to solve derivatives, you'll find yourself doing the same manipulations over and over again for different problems.

Instead of wasting time like this, we can find efficient methods to differentiate (find the derivative of) new functions using differentiations that we already know.

For example, if we have a function h(x)=f(x)g(x)h(x) ={f(x)}{g(x)} and we want to find

dh(x)dx,\frac{d h(x)}{dx},

then we can find a formula for this in terms of f(x)f(x) and g(x)g(x), since h(x)h(x) is defined using them.

Try it on your own first, and plug it in when you think you've got it.

df(x)g(x)dx=\displaystyle{\frac{df(x)g(x)}{dx}}=

Hint 1

Can't distribute?

abcdab-cd
=abad+adcd=ab-ad+ad-cd
=a(bd)+d(ac).=a(b-d)+d(a-c).

Yes you can.

Hint 2

What is

limδx0f(x+δx)f(x)δx=??\lim_{\delta x \to 0} \frac{f(x+\delta x)-f(x)}{\delta x}=??
Solution 1
  1. Set up the limit:
limδx0f(x+δx)g(x+δx)f(x)g(x)δx\lim_{\delta x \to 0} \frac{f(x+\delta x)g(x+\delta x) -f(x)g(x)}{\delta x}
  1. Apply the factorization rule (see Hint 1):
limδx0f(x+δx)g(x+δx)f(x+δx)g(x)+f(x+δx)g(x)f(x)g(x)δx\lim_{\delta x \to 0} \frac{f(x+\delta x)g(x+\delta x)-f(x+\delta x)g(x)+f(x+\delta x)g(x) -f(x)g(x)}{\delta x}
  1. Separate:
limδx0f(x+δx)g(x+δx)f(x+δx)g(x)δx+limδx0f(x+δx)g(x)f(x)g(x)δx\lim_{\delta x \to 0} \frac{f(x+\delta x)g(x+\delta x)-f(x+\delta x)g(x)}{\delta x}+ \lim_{\delta x \to 0}\frac{f(x+\delta x)g(x) -f(x)g(x)}{\delta x}
  1. Factor:
limδx0f(x+δx)limδx0g(x+δx)g(x)δx+g(x)limδx0f(x+δx)f(x)δx\lim_{\delta x \to 0}f(x+\delta x)\cdot\lim_{\delta x \to 0} \frac{g(x+\delta x)-g(x)}{\delta x}+ g(x)\lim_{\delta x \to 0}\frac{f(x+\delta x) -f(x)}{\delta x}
  1. These are just differentiations of f(x)f(x) and g(x)g(x)! So, we can

substitute their differentiations:

limδx0f(x+δx)g(x)+g(x)f(x)\lim_{\delta x \to 0}f(x+\delta x)\cdot g'(x)+ g(x)f'(x)
  1. Finally, we approximate:
δx0\delta x \rightarrow 0
df(x)g(x)dx=f(x)g(x)+g(x)f(x)\frac{df(x)g(x)}{dx}=\boxed{ f(x)g'(x)+ g(x)f'(x)}

That is what we call the Product Rule.

Now division!

df(x)g(x)dx=\displaystyle{\frac{d\frac{f(x)}{g(x)}}{dx}}=

Hint 1

Remember the distribution rule from the previous problem?

Hint 2

Make two huge fractions, simplify, and add them.

Solution 1
  1. Set up the limit:
limδx0f(x+δx)g(x+δx)f(x)g(x)δx\lim_{\delta x \to 0} \frac{\frac{f(x+\delta x)}{g(x+\delta x)} -\frac{f(x)}{g(x)}}{\delta x}
  1. Apply the factorization rule (see Hint 1 from previous problem):
limδx0f(x+δx)g(x+δx)f(x+δx)g(x)+f(x+δx)g(x)f(x)g(x)δx\lim_{\delta x \to 0} \frac{\frac{f(x+\delta x)}{g(x+\delta x)} -\frac{f(x+\delta x)}{g(x)}+\frac{f(x+\delta x)}{g(x)} -\frac{f(x)}{g(x)}}{\delta x}
  1. Separate:
limδx0f(x+δx)g(x+δx)f(x+δx)g(x)δx+f(x+δx)g(x)f(x)g(x)δx\lim_{\delta x \to 0} \frac{\frac{f(x+\delta x)}{g(x+\delta x)} -\frac{f(x+\delta x)}{g(x)}}{\delta x}+\frac{\frac{f(x+\delta x)}{g(x)} -\frac{f(x)}{g(x)}}{\delta x}
  1. Combine:
limδx0f(x+δx)g(x)g(x+δx)g(x)g(x+δx)δx+1g(x)f(x+δx)f(x)δx\lim_{\delta x \to 0}\frac{f(x+\delta x)}{g(x)g(x+\delta x)}\cdot\frac{g(x) -g(x+\delta x)}{\delta x}+\frac{1}{g(x)}\cdot\frac{f(x+\delta x)-f(x)}{\delta x}
  1. These are just differentiations of f(x)f(x) and g(x)g(x)! So, we can

substitute their differentiations:

limδx0f(x+δx)g(x)g(x+δx)(g(x))+1g(x)f(x)\lim_{\delta x \to 0}\frac{f(x+\delta x)}{g(x)g(x+\delta x)}(-g'(x))+\frac{1}{g(x)}f'(x)
  1. Now, we approximate:
δx0\delta x \rightarrow 0
f(x)g(x)g(x)g(x)+f(x)g(x)\frac{f(x)}{g(x)g(x)}g'(x)+\frac{f'(x)}{g(x)}
=f(x)g(x)g(x)f(x)(g(x))2= \boxed{\frac{f'(x)g(x)-g'(x)f(x)}{(g(x))^2}}

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