Techniques for Finding Derivatives

Techniques of Differentiation

The definition of a derivative, $f'(x)=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}$ is not the only method of finding a derivative. This section provides three different techniques to help find derivatives. These techniques are often quicker than the definition of a derivative, depending on the given equation. The last part of the section presents how to calculate higher order derivatives.

The Power Rule

The first technique allows you to find a derivative of a summation of multiple terms.

This is known as The Power Rule.

Power Rule only applies in certain conditions.

Theorem

If n is a rational number, then the function $f(x)=x^{n}$ is differentiable and

${\frac {d}{dx}}[x^{n}]=nx^{n-1}.$

For f to be differentiable at x=0, n must be a number such that $x^{n-1}$ is defined on an interval containing 0.

Proof

${\frac {d}{dx}}[x^{n}]=\lim _{{\Delta }x\rightarrow 0}{\frac {(x+{\Delta }x)^{n}-x^{n}}{{\Delta }x}}$

$=\lim _{{\Delta }x\rightarrow 0}{\frac {x^{n}+nx^{n-1}({\Delta }x)+{\frac {n(n-1)x^{n-2}}{2}}({\Delta }x)^{2}+.....+({\Delta }x)^{n}-x^{n}}{{\Delta }x}}$

$=\lim _{{\Delta }x\rightarrow 0}\left[nx^{n-1}+{\frac {n(n-1)x^{n-2}}{2}}({\Delta }x)+.....+({\Delta }x)^{n-1}\right]$

$=nx^{n-1}+0+.....+0$

$=nx^{n-1}{\square }.$

Example

The following is an example of the Power Rule:

$f(x)=3x^{2}+2x+1$

Normally you would work through the definition of a derivative,

$f'(x)=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}$

but instead you can apply the power rule to find your derivative.

First, take a look at the first term of the equation, $3x^{2}.$

To find $f'(x)$ first drop the power on the variable of this term down to the front of the equation, $(2)*(3x^{2}).$

Next, subtract one from the power that was dropped down in front of the equation, $(2)*(3x^{1}).$

Or, $(2)(3)x=6x.$ So now we have the first part of the derivative, $6x.$

Now, perform this operation to the second and the third terms of the equation.

For the second part, $2x^{1}$

$(1)*(2x^{1})$

$(1)(2)(x^{0})=2.$

This shows that the second part is $2.$

Now for the third part we notice that there is no variable, $x.$

There are two ways to look at this last part, either: one, there is no variable so the part disappears, or two, since there is no variable, we multiply the whole piece by 0, since that would be the power of the variable.

Either way, there is no longer a third piece.

Now, if we put the two parts together, we have our answer

$f'(x)=6x+2.$

Here is another example of this method in action

$f(x)=3x^{3}-4x^{2}+3x-4$

$f'(x)=(3)(3x^{3-1})-(2)(4x^{2-1})+3x^{1-1}$

$f'(x)=9x^{2}-8x+3.$

Here are some practice problems.

$f(x)=x^{2}-7x+13$

$h(x)=2x^{3}+2x-3$

$g(x)=4x^{4}-7x^{3}+2x^{2}+5x-4$

The Product Rule

The next technique is called The Product Rule.

As previously stated, not all derivatives can be found through the power rule.

While it is a nice way to find many derivatives, other methods exist depending on the equation you are currently solving.

The Power Rule can be seen as the sum of two separate functions, and their respective derivative is the sum of the individual derivatives.

The product rule introduces how to solve the product of two functions.

Theorem

The product of two differentiable functions $f(x)$ and $g(x)$ is itself differentiable.

The derivative of $f(x)g(x)$ is $f(x)*g'(x)+g(x)*f'(x).$

Proof

${\frac {d}{dx}}[f(x)g(x)]=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)*g(x+{\Delta }x)-f(x)g(x)}{{\Delta }x}}$

$=\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)*g(x+{\Delta }x)-f(x+{\Delta }x)g(x)+f(x+{\Delta }x)g(x)-f(x)g(x)}{{\Delta }x}}$

$=\lim _{{\Delta }x\rightarrow 0}\left[f(x+{\Delta }x)*{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}+g(x)*{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}\right]$

$=\lim _{{\Delta }x\rightarrow 0}\left[f(x+{\Delta }x)*{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}\right]+\lim _{{\Delta }x\rightarrow 0}\left[g(x)*{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}\right]$

$=\lim _{{\Delta }x\rightarrow 0}f(x+{\Delta }x)*\lim _{{\Delta }x\rightarrow 0}{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}+\lim _{{\Delta }x\rightarrow 0}g(x)*\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}$

$=f(x)g'(x)+g(x)f'(x){\square }.$

Example

For example, take the equation $h(x)=(x^{2}+1)(x^{2}-2x-1).$

To start, think of the derivative of each part. Let $f(x)=x^{2}+1$ and let $g(x)=x^{2}-2x-1.$

$f'(x)=2x$ and $g'(x)=2x-2.$

Now apply each part to the previous theorem.

$h'(x)=f(x)*g'(x)+g(x)*f'(x)$

$h'(x)=(x^{2}+1)*(2x-2)+(x^{2}-2x-1)*(2x).$

Now all we need to do is simplify.

$h'(x)=(2x^{3}+2x-2x^{2}-2)+(2x^{3}-4x^{2}-2x)$

$h'(x)=(2x^{3}-2x^{2}+2x-2)+(2x^{3}-4x^{2}-2x)$

$h'(x)=4x^{3}-6x^{2}-2.$

Now, if product rule doesn't seem to appeal to you, when possible, you can just simplify the original equation.

If you multiple the parts by each other and then find the derivative, you are simply doing the Power Rule.

Sometimes this may be easier, but in some cases you may be unable to multiply the parts together.

Here is an example of changing a product rule.

$h(x)=(x^{2}+1)(x^{2}-2x-1)$

$h(x)=x^{4}+x^{2}-2x^{3}-2x-x^{2}-1$

$h(x)=x^{4}-2x^{3}-2x-1.$

Then using the power rule,

$h'(x)=4x^{3}-6x^{2}-2.$

As you can see, both methods yield the same answer.

Also, product rule can be expanded beyond two pieces.

This would look like:

${\frac {d}{dx}}[f(x)g(x)h(x)]=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x).$

For further work, here are some practice problems.

$h(x)=(x^{2}-2)(3x+1)$

$w(x)=(3x+2)(2x^{2}+4x)$

$p(y)=(y^{2}-3y)(5y-4)(2y^{2}+1)$

The Quotient Rule

The next technique is called The Quotient Rule .

Before we learned what to do with the product of two functions, quotient rule teaches us what to do with a function divided by another function.

Theorem

The quotient of two differentiable functions, ${\frac {f(x)}{g(x)}}$ is itself differentiable for all $g(x)\neq 0.$

This derivative is given by the bottom function multiplied by the derivative of the top function, minus the top function multiplied by the derivative of the bottom function, all divided by the bottom function squared.

Or, ${\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]={\frac {g(x)f'(x)-f(x)g'(x)}{[g(x)]^{2}}}.$

Proof

${\frac {d}{dx}}\left[{\frac {f(x)}{g(x)}}\right]=\lim _{{\Delta }x\rightarrow 0}{\frac {{\frac {f(x+{\Delta }x)}{g(x+{\Delta }x)}}-{\frac {f(x)}{g(x)}}}{{\Delta }x}}$

$=\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)f(x+{\Delta }x)-f(x)g(x+{\Delta }x)}{{\Delta }xg(x)g(x+{\Delta }x)}}$

$=\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)f(x+{\Delta }x)-f(x)g(x)+f(x)g(x)-f(x)g(x+{\Delta }x)}{{\Delta }xg(x)g(x+{\Delta }x)}}$

$={\frac {\lim _{{\Delta }x\rightarrow 0}{\frac {g(x)[f(x+{\Delta }x)-f(x)]}{{\Delta }x}}-\lim _{{\Delta }x\rightarrow 0}{\frac {f(x)[g(x+{\Delta }x)-g(x)]}{{\Delta }x}}}{\lim _{{\Delta }x\rightarrow 0}[g(x)g(x+{\Delta }x)]}}$

$={\frac {g(x)[\lim _{{\Delta }x\rightarrow 0}{\frac {f(x+{\Delta }x)-f(x)}{{\Delta }x}}]-f(x)[\lim _{{\Delta }x\rightarrow 0}{\frac {g(x+{\Delta }x)-g(x)}{{\Delta }x}}]}{\lim _{{\Delta }x\rightarrow 0}[g(x)g(x+{\Delta }x)]}}$