## What is Implicit Differentiation?

Usually when you start to take derivatives you begin with equations that say

y = ……..

You see y on the left side of the equation, but only once and by itself. And at first there is generally just one variable on the right side of the equation.

You can however find the derivative of an equation if y is not isolated and if y has other operations being done upon it. The variable y may also show up more than once. The chain rule is required.

Implicit Differentiation

Again, you definitely want to remember the chain rule when you do implicit differentiation!

Later on you will solve for dy/dx or dz/dx or whichever term you are trying to find.

Example:

dy/dx appears both on the left side and right side of the equation. You can combine them since they are similar.

## Anti-Derivative/Integral of sin(2x) Using u-substitution

The anti derivative of sin(2x) means something that when you take the derivative of it will be sin(2x). Indicated by the empty box here.

You can also think about it as the indefinite integral.

A technique called u-substitution is useful here. I choose u = 2x

By taking the derivative of u, I can get it in terms of dx which I can then substitute back into the integral. Then the variable in the integral is u everywhere. At the end we’ll substitute back for u, since the definition of u is decided by us.

You add the constant at the end since there could be a constant added without changing the derivative. The derivative of constants is zero.

## How to Take a Derivative with a Square Root

Rewrite the square root as what is inside the parentheses to the 1/2 power.

Then you do the usual process of multiplying by the exponent, bringing the 1/2 power exponent down by one, to become -1/2 and then doing the chain rule.

## Solving a Somewhat Tricky Limit Problem

This is an interesting problem.

What I would do first is try plugging in a somewhat large number, like 1000 for x.

It’s not infinity, but it’s large.

I removed the interior sets of parentheses since parentheses are implied in a fraction like that.

The expression then becomes a number a bit larger than 1, to an exponent of 2000.

If you go ahead and use a calculator, it becomes

That seems distinctly non zero to me!

So I tried a bigger number, x=1,000,000

You get

It’s not immediately apparent what that value is. (you might try taking the natural log of this number though)

An approach which uses an early definition of e can help you solve this problem.

There is another approach you could use. One of the first steps is to put a base of e with the natural log of the expression. And we will need to use a Taylor Expansion later.

You may have seen the approximation that sinx ~ x for small values, there is a similar approximation for ln(1+x), 5/infinity qualifies as a small value. The first term of the Taylor Expansion is the most dominant for small values.

## Steps for Finding a Derivative of a Function with a Square Root

My process would be to first rewrite the square root by using a fractional exponent (1/2).

From there, you multiply by the exponent in front and then subtract one from the exponent. Then you do the chain rule.

You can rewrite the answer using square roots if you would like.

## Tutoring Calculus, Clock minute hand area problem and the natural number e

We started by looking at one of the more difficult problems on her assignment that had to do with the changing area of the section swept out by the minute hand of the clock. Her teacher had made the simplification of disregarding the movement of the hour hand (but I think that to be more accurate, you would want to take that into account since I think on most clocks the hour hand does move slowly as the minutes progress, it would change the result by about 2%).

The approach for a problem like that is to figure out what things change, the variable, and what is constant. At first she thought that the radius of the minute hand was a variable, but since it does not change in this problem, it is also constant, just like numbers. Using the variable theta to describe the angle swept out by the minute hand helped solve the problem, though that variable was not mentioned explicitly in the directions.

She was a bit unfamiliar with the natural number e. It functions like any other constant in most of the problems and is related to the natural log function.

## Tutoring Calculus, Mean Value Theorem and Optimization

We started by looking at some problems with derivatives, they included the product rule, quotient rule, and the chain rule. The format of the problems was a bit different than usual.

After that, we looked at the mean value theorem and what it really means. Given two points and the values of a function in between, if the function is differentiable than one (possibly more) point will have the same slope as the average slope. The average slope is found using methods used in algebra.

Also looked at some optimization problems where the basic approach is to find an equation and then set the first derivative equal to zero since the slope will be zero at maxima and minima (and inflection points).

## Tutoring Calculus, Chain Rule

We started by looking at extrema for functions, maxima and minima. They can be seen on graphs with a flat slope and can also be calculated by using the first derivative.

We looked at derivatives of some trigonometric functions including arc sine as well as the domain of the functions. It can help to look at the functions in a few ways. Two ways are on a unit circle and also as an oscillating function on the x-y graph.

U-substitution is a useful approach and includes the chain rule. I actually have a table of derivatives that uses u each time.  That can be a good way to think about derivatives since many times there will be something other than x that is being operated on by the function.