Each of you has much information on finding derivative formulas using theorems from previous course work. The following are ideas that will make the " paper and pencil " computations of partial derivatives relatively easy. These ideas can be thought of as review, as summaries, and/or as " other ways to look at " some old ideas.

1. We needn't feel dependent on formula sheets for many simple computations :

By " talking " the formula we get to " know " it.Many examples follow.

The derivative of a constant multiple is the constant factor times the result of the derivative.

f(x) = 5sin(x) then f ' (x) = 5cos(x)

The derivative of a sum is the dum of the derivatives.

The derivative of a product is the sum of products of the functions and their derivatives, mixed not matched .

g(x) = sin(x) ln(x) then g ' (x) = cos(x) ln(x) + (1/x) sin(x) If there were more factors there would be more terms in the result but each would contain the same number of factors as the original problem and each contains exactly one derivative until all are exhausted. For example p(x) = [cos(x)][ sqrt(x) ][ln(x)]

then p ' (x) = [sin (x)][ sqrt(x)][ ln(x)] + [cos(x)][ (1/ 2 sqrt(x))][ ln(x)] + [cos(x)][sqrt(x)][(1/x)]

[ Did we forget that the derivative of the ln(x) is the reciprocal of x , that is , 1/x .]"Talk" them out.

Notice the different thought on sqrt(x) . You learned to write x raised to the one-half power each time you differentiated it. Then you used a theorem you called a "power rule". Instead, " talk " the result. The derivative of the square root function is the reciprocal of the double. 

And on the matter of the " power rule ", try instead : the derivative of a polynomial is a polynomial of one less degree. You will need to recognize the new coefficient.Put the emphasis where it will help.

For Trig. Functions : 

Sine and Cosine are said to be cofunctional ( it's in the name ). The other four are defined in pairs that are cofunctional.[ There may be a reason for this---it makes it easier to remember theorems. ]

Notice the results after differentiation are also cofunctional but with a sign change. Remember the one withoutthe minus sign ; then the other must have the minus sign.

The exponential function is " outstanding ".  It is it's own derivative. 

And why would we use Exp(x) instead of our "old friend" 

Think about this: You are able to find the name of the function in Exp(x) but you are not able to find it in the expression 

It may have been for this reason that you struggled with functional composition when the exponential function was present.

And how does differentiation " work " on compositions. "Talk" it out: The derivative of a composition is a productof the derivatives of the functions involved and the arguments in those functions are the same as in the problem.

If p(x)= sin(ln(sqrt(x))) then p ' (x) = cos[ ln(sqrt(x)][ 1/(sqrt(x)][ 1/(2sqrt(x)]

Again, the derivative of ln(x) is the reciprocal of x

The argument is the item in the " window ". For sin(ln(sqrt(x))) , sqrt(x) is the argument for ln and

ln(sqrt(x)) is the argument for sin . Yes, and x is the argument for sqrt and it's derivative is 1 and therefore is not "visible" in the result of the differentiation.

The derivative of a quotient can be understood after looking at a special case, namely, the derivative of the reciprocal of a function is the quotient of that functions derivative divided by the square of that function and there is a sign change. Now that was a mouth full. Let's see : 

then the theorem for the quotient of two functions can be seen as a combination of the previous result and the product theorem. [Why is the product theorem relevant in a quotient problem ?]

Now we are ready to compute partial derivative formulas for functions of two independent variables.

See 13.3 1st item Partial Derivatives

See 13.3 Example 1 Finding Partials

See 13.3 Example 2 Finding and Evaluating Partials

See 13.3 Example 3 Slopes

See 13.3 Example 5 Rates

See 1.3 Example 6 Finding Partials

See 13.3  Higher Order Partials

See 13.3 Example 7 Finding Higher Order Partials

Look at the following solved problems in the Exercises at the end of 13.3 : 5-21odd, 29, 31,35, 43, 47, 59, 60 Then do the following unsolved problems as homework: 6-22even, 30, 32, 44, 48, 50, 60