More About Derivatives (Part 2)


Notation: The derivative f(x)f'(x) of a function f(x)f(x) is also denoted as

dfdx(x)\frac{df}{dx} (x)

This notation comes from the fact that when you compute the derivative, you compute

dfdx(x0)=limxx0f(x)f(x0)xx0\frac{df}{dx}(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}

The term f(x)f(x0)f(x)-f(x_0) is usually referred as Δf\Delta f, and the term xx0x-x_0 is referred as Δx\Delta x. So, sometimes, in some books (specially Physics books) you're going to find the definition

dfdx(x0)=limΔx0ΔfΔx\frac{df}{dx}(x_0) = \displaystyle\lim_{\Delta x\to 0} \frac{\Delta f}{\Delta x} Theorems to calculate Derivatives

Now it is the time to introduce the heavy artillery. In practice, you won't be computing the limit

f(x0)=limxx0f(x)f(x0)xx0f'(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}

very often. It is very important to know how to do it that way, but most of the times it won't be necessary.

Examples of Derivatives

Example: Compute the derivative of the function f(x)=x3+x2f(x)=x^3+x^2.

Solution: What do we do here, do we apply the limit to compute the derivative?? Well, your line of reasoning should be the following: The function f(x)=x3+x2f(x) = x^3+x^2 corresponds to the sum of x3x^3 and x2x^2. The intuition is that if I could compute the derivative of each term separately, then I could simplify the calculation.

In other words, if I knew what is the derivative of x3x^3, and if I also knew what is the derivative of x2x^2, then I should know what is the derivative of x3+x2x^3+x^2.....

\star In fact, you do. We have the following theorem:

Theorem: The Derivative of the Sum of Two Functions

Assume that f(x)f(x) and g(x)g(x) are differentiable at x0x_0 (that means that the derivative exists at that point). Then, we have that

ddx(f(x)+g(x))=dfdx(x)+dgdx(x) \frac{d}{dx}(f(x)+g(x)) = \frac{df}{dx}(x) + \frac{dg}{dx}(x)

In other words, the derivative of the sum is the sum of the derivatives (These are not empty words, they really describe the result accurately). This is usually referred as the Linearity Property of the derivative

Now we show a result which is one of the most important derivative rules, that will help us to compute a lot of derivatives:

Theorem: The following holds true for all n0n\ne 0:

ddxxn=nxn1 \frac{d}{dx} x^n = n x^{n-1}

Proof: We won't do anything too deep, just to save you from fatal boredom, but let's just do this one to get a feel of it. By definition

dfdx(x0)=limxx0f(x)f(x0)xx0=limxx0xnx0nxx0\frac{df}{dx}(x_0) = \displaystyle\lim_{x\to x_0} \frac{f(x)-f(x_0)}{x-x_0}=\displaystyle\lim_{x\to x_0} \frac{x^n- x_0^n}{x-x_0} =limxx0(xx0)(xn1+xn2x0+...+xx0n2+x0n1)xx0=limxx0(xn+xn1x0+...+xx0n+x0n)= \displaystyle\lim_{x\to x_0} \frac{(x- x_0)(x^{n-1}+x^{n-2}x_0 + ...+ x x_0^{n-2} +x_0^{n-1})}{x-x_0} = \displaystyle\lim_{x\to x_0}(x^n+x^{n-1}x_0 + ...+ xx_0^n +x_0^n) =x0n1+x0n2x0+...+x0x0n2+x0n1=nx0n1 = x_0^{n-1}+x_0^{n-2}x_0 + ...+ x_0x_0^{n-2} +x_0^{n-1} = n x_0^{n-1}

So, let's come back to the problem of finding the derivative of f(x)=x3+x2f(x) = x^3+x^2. Using the Linearity of derivatives we find that

ddx(x3+x2)=ddxx3+ddxx2=3x2+2x\frac{d}{dx}(x^3+x^2) = \frac{d}{dx} x^3 + \frac{d}{dx}x^2 = 3x^2 + 2x

Let's recall that ddxxn=nxn1\frac{d}{dx}x^n = nx^{n-1}, so applying that to the case n=3n=3 and n=2n=2 respectively we get the previous result. The Linearity property can be written in a more general way:

Theorem: Assume that f(x)f(x) and g(x)g(x) are differentiable at x0x_0 and aa and bb are constants. Then

ddx(af(x)+bg(x))=adfdx(x)+bdgdx(x) \frac{d}{dx}(af(x)+ bg(x)) = a\frac{df}{dx}(x) + b\frac{dg}{dx}(x)

Below we show an example of how to apply this result:

Example: Compute the derivative of the function f(x)=3x3+2x1/2f(x)=3 x^3+2 x^{1/2}.

Solution: Using Linearity, we get that

ddx(3x3+2x1/2)=3ddxx3+2ddxx1/2=3(3x2)+2(12x1/2)\frac{d}{dx}(3x^3+ 2x^{1/2}) = 3\frac{d}{dx} x^3 + 2 \frac{d}{dx}x^{1/2} = 3(3x^2) + 2\left(\frac{1}{2} x^{-1/2}\right) =9x2+x1/2 =9x^2 + x^{-1/2}

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