Power Rule and Implicit Differentiation

This is the follow up of my calculus post. This post investigates one of the rules of calculus that calculates the derivative without using the painfully long equation of the derivative.

Power Rule

Now, let’s start with the derivative:

\frac{\delta x}{\delta y}=\frac{f(x+\delta x)-f(x)}{\delta x} The limit of delta x goes to 0, remember that.

If the function is nx, and plug it in, then the result is n, as shown in my first post. If it is nx^2, then the derivative is 2nx. I wonder what the result is for nx^3?

\frac{\delta x}{\delta y}=\frac{(x+\delta x)^3-x^3}{\delta x} Multiply the polinomials (you know how, right?)

\frac{\delta x}{\delta y}=\frac{x^3+3x^2\delta x+3x\delta x^2+\delta x^3-x^3}{\delta x} Scratch out x^3

\frac{\delta x}{\delta y}=\frac{3x^2\delta x+3x\delta x^2+\delta x^3}{\delta x} Divide by delta x

\frac{\delta x}{\delta y}=3x^2+3x\delta x+\delta x^2 Now. the limit of delta x goes to zero, therefore:

\frac{\delta x}{\delta y}=3x^2

So, the derivative of X^3 is 3x^2. So far, so good. For x, it is 1, for x^2, it is 2x, and I mentioned the x^3 one… Hold on, now, do you notice a pattern for the derivatives? In every single one of them, the number of the exponent is multiplied into the beggining, and the exponent substracts by one! Therefore, the general rule for the derivative of x to the power of blah, I suspect, is this:

Cnx^{n-1} (C is any constant)

Now, how do I know it is true? Well, let’s prove it! (You can skip this one to the implicit differentiation, since the proof is just icing on the cake)I will prove it for all integers (-2,-1,0,1,2, etc…) using the binomial theorem, which is a formula that is true for all cases of binomial multiplication. (Oh boy, it is going to be monstrous)So, here goes nothing:

\frac{\delta x}{\delta y}=\frac{f(x+\delta x)-f(x)}{\delta x} Insert the binomial theorem:

\frac{\delta x}{\delta y}=\frac{x^n+\frac{n!}{(n-1)!1!}x^{n-1}\delta x+\frac{n!}{(n-2)!2!}x^{n-2}\delta x^2}{\delta x}+…

…+\frac{\frac{n!}{(n-(n-1))!(n-1)!}x\delta x^{n-1}+\delta x^n-x^n}{\delta x} Phew, that was a mouthful! Now, let’s simplify… Ok, x^n adds to -x^n, cancelling both terms. Next, the second term and the fourth term is simplified to n because:

n(n-1)(n-2)…(1)/(n-1)(n-2)…(1)=n (no extra n in the bottom one because it is (n-1)!)

\frac{\delta x}{\delta y}=\frac{nx^{n-1}\delta x+\frac{n!}{(n-2)!2!}x^{n-2}\delta x^2}{\delta x}+…

…+\frac{nx\delta x^{n-1}+\delta x^n}{\delta x} Divide by delta x, and distribute a constant inside.

\frac{\delta x}{\delta y}=Cnx^{n-1}+C\frac{n!}{(n-2)!2!}x^{n-2}\delta x+…

…+Cnx\delta x^{n-2}+C\delta x^{n-1}

Now, since we take the limit of delta x is 0, so \delta x goes smaller, \delta x, smaller, \delta x, smaller, \delta x, smaller, \delta x until BANG! It blows up from all the air getting sucked out of it.

Therefore, the derivative of any monomial to the nth power is *drum rolls* \frac{\delta x}{\delta y}=Cnx^{n-1}

Yay! My original assumption was true!

Implicit Differentiation

Now, we are going to use the previous assumption about the power rule of monomials with exponents of integers to prove the power rule for all monomials with rational exponent and teach implicit differentiation. By the way, I will not prove for all real numbers because I don’t know how, K? Now, implicit differentiation is basically differentiating two or more variables, like y=x. For, say, x^2+x, you differentiate the individual parts, but for y=x, you do something a little bit different. Now, let’s prove the monomial with rational exponent:

y=x^{\frac{p}{q}} p/q represents a rational number, which is any number that can be represented by a fraction. Now, to differentiate, firstly, the y must have an exponent, so multiply both sides by q on the exponent:

y^q=x^{\frac{pq}{q}} This makes both sides monomials with integer exponents. y^q=x^p When the power rule is used:

qy^{q-1}\frac{\delta y}{\delta x}=px^{p-1} In an implicit differentiation, you write in another variable the derivative AND differentiate both variables. Since you are trying to find the derivative, you isolate the derivative, and move the other parts to the other side:

\frac{qy^{q-1}}{qy^{q-1}}\frac{\delta y}{\delta x}=\frac{px^{p-1}}{qy^{q-1}} Goes to: \frac{\delta y}{\delta x}=\frac{px^{p-1}}{qy^{q-1}}

Uh oh, this smells like trouble. How can you divide the exponents into a simpler term? Well, you turn the function y into x. Remember, y=x^(p/q):

\frac{\delta y}{\delta x}=\frac{p}{q}\frac{x^{p-1}}{x^{\frac{p}{q}(q-1)}} Distribute the bottom fraction in the exponent:

\frac{\delta y}{\delta x}=\frac{p}{q}\frac{x^{p-1}}{x^{\frac{pq}{q}-\frac{p}{q}}}

\frac{\delta y}{\delta x}=\frac{p}{q}\frac{x^{p-1}}{x^{p-\frac{p}{q}}}
\frac{\delta y}{\delta x}=\frac{p}{q}\frac{x^{p-1}}{x^{\frac{pq}{q}-\frac{p}{q}}} Substract the top and bottom of the fraction by the bottom.
\frac{\delta y}{\delta x}=\frac{p}{q}x^{p-1-(p-\frac{p}{q})} Distribute and collect to get:

\frac{\delta y}{\delta x}=\frac{p}{q}x^{\frac{p}{q}-1} This makes the power rule true! Plus, you learned something about implicit differentiation, and a few algebraic skills during the reading.

Wow, I am finally finished. Anyways, as you see with the power rule, you don’t have to use the derivative all the time, and this makes life easier. Not only that, the coolest things of these rules, and the whole of calculus, is that you can find the slope of a curve at any point, even if x equals some crazy number, and using some of these techniques is the key to figuring them out.


One Response to Power Rule and Implicit Differentiation

  1. […] to Calculus By ibyea I made two posts regarding that. All other calculus stuff, you can look it up in visual calculus, a very good page […]

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