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22M:017, Fall 2006 Lecture 13 (9/20/06)
Juan A. Gatica
We have seen the Chain Rule. We will spend a great deal of time today doing examples, since this is an important topic.
Example. The limit
lim h 0
2 h^3 1 3 h is of the form fx 0 . Find f, x 0 and the limit. Solution. Recall that
fx 0 lim h 0 fx^0 ^ h^ ^ fx^0 h
Thus we must have fx x^3 1 x 0 2. Then fx 1 2 x^3 1
3 x^2 (using the Chain Rule).
Then
lim h 0
2 h^3 1 3 h f 2
(^126) 2. Example. Do the same as in the previous example for lim h 0 e 1 h^2 1 h (^) e 2 h
With the same reasoning: fx e x^2 x, x 0 1. Then
fx 2 x 1 e x^2 x f 1 3 e^2. Example. As above, find
lim h 0 log 3 3 h^2 3 h log 3 12 h
Here fx log 3 x^2 x, x 0 3. To find the derivative of log 3 x^2 x we MUST change the base in order to use the natural logarithm. We also must use the chain rule.
fx log 3 x^3 x 1 ln 3 lnx^3 x
Then: fx 3 x
ln 3 x^3 x f 3 28 30 ln 3
Thus:
lim h 0 log 3 3 h^2 3 h log 3 12 h ^
30 ln 3
Example. Find the derivative of the function fx e x^ lnx^2 1 ^10. In this example, if we let hx e x^ lnx^2 1 , gx x^10 then fx g hx, so we can use the chain rule. Since hx e x^ (^) x^22 x 1 , gx 10 x^9 we get:
fx ghxhx 10 e x^ lnx^2 1 ^9 e x^ 2 x x^2 1
In general if fx is a differentiable function and r is a real number and if gx fxr^ is defined, then g is differentiable and:
gx rfxr^1 fx.
Example. Find the slope of the line tangent to the graph of fx x^2 x 1 23 at the point 1, 3 (^23) . The slope is f 1 and fx 23 x^2 x 1 ^ (^13) 2 x 1 so:
fx e^
x (^) exe x (^) ex e x (^) exe x (^) ex e x^ ex^2
e^
x (^) ex (^2) e x (^) ex 2 e x^ ex^2
e^
x (^) ex (^) e x (^) exe x (^) ex (^) e x (^) ex e x^ ex^2
2 e x 2 ex e x^ ex^2 4 e x^ ex^2
Example. Find the derivative of the function fx x^2. Here we have to use the chain rule and change the base of the exponential:
fx eln x^2 fx 2 x ln eln x^2 2 x ln x^2.
