The Table Method for Derivatives

1. Math 170 Problem Examples -L'Hopital's Rule, Complicated Derivatives Tyler Murphy October 14, 2014 Here are two worked problems from the homework. Recall my method for solving problems in calculus: First, identify the situation (chain rule, product rule, etc.). Second, make a table of f; f0; g; g0. Third, plug values from table into the formula determined by the situation. Fourth, simplify. 1 Find the Derivative of y = ln x2 This is a chain rule problem with: f = x2 f0 = 2x g = ln x g0 = 1 x So we put this in the chain rule: [f(g(x))]0 = f0(g(x)) g0(x) We get: 2(ln x) 1 x = 2 ln x x 1

2. 2 Find the limit: limx!0 sin 2x sin 7x Putting 0 into the equation gives us 0 0 so we can use L'Hopital's Rule. f = sin 2x f0 = 2 cos 2x g = sin 7x g0 = 7 cos 7x So we have: lim x!0 2 cos 2x 7 cos 7x Pluging in 0, we get: lim x!0 2 cos (2 0) 7 cos (7 0) = 2 cos (0) 7 cos (0) = 2 1 7 1 = 2 7 3 Find the derivative of cos (tan (ln ( x3 3 ex))) This problem looks horrible. We have a product rule inside a chain rule inside a chain rule inside a chain rule. But let's start at the outside and work inwards, one piece at a time, using the tables outlined in my method at the beginning of this paper. f = cos x f0 = sin x g = tan (ln ( x3 3 ex)) g0 = ? Calculating g0 will be dicult as is, so let's apply the table method again. To avoid confusion, I'll use h instead of f and j instead of g. Note that g0 is a product rule inside a chain rule inside a chain rule. h = tan x h0 = sec2 x j = ln ( x3 3 ex) j0 = ? Now we do it again to

3. nd j0. Note that j0 is a product rule inside a chain rule. 2

4. m = ln x m0 = 1 x n = x3 3 ex n0 = ? Now we do it one last time to

5. nd n0, which is a product rule. p = x3 3 p0 = x2 q = ex q0 = ex Since this table it complete, we can use it to build the product rule, which is n0 from the previous table. n0 = p0q + q0p = (x2)(ex) + (ex)( x3 3 ) = (ex)(x2 + x3 3 ) we now put this value into the m; n table and use it to back

6. ll the previous table. m = ln x m0 = 1 x n = x3 3 ex n0 = (ex)(x2 + x3 3 ) We can now use this table to

7. nd j0 which is a chain rule for m(n(x)) j0 = m0(n(x))n0 = 2 664 1 x3 3 (ex) (ex)(x2 + x3 3 ) 3 775= 3 x3 (x2 + x3 3 ) = 3x2 x3 + 3 x3 x3 3 = 3 x + 1 So now we have j0 which we can put into the h; j table: h = tan x h0 = sec2 x j = ln ( x3 3 ex) j0 = 3 x + 1 Now recall that we made the h; j table to

8. nd g0 in our

9. rst table, which was a chain rule of h(j(x)). 3

10. So. g0 = h0(j(x)) j0(x) = sec2 (ln ( x3 3 ex)) ( 3 x + 1) Now we can plug this value into our f; g table. We could attempt to simplify this

11. rst, but I don't think we can. f = cos x f0 = sin x g = tan (ln ( x3 3 ex)) g0 = sec2 (ln ( x3 3 ex)) ( 3 x + 1) Now we

12. nally arrive at the

13. rst table we made for the original derivative, which was a chain rule of f(g(x)). So we have: f0(g(x)) g0(x) = sin (tan (ln ( x3 3 ex))) (sec2 (ln ( x3 3 ex)) ( 3 x + 1)) That was a long problem. But when we took it piece by piece, and dealt with each situation on its own, it wasn't a dicult problem. Note that there is a dierence between a dicult problem and a long problem. This is why I prefer the table method. We can take a problem that would be both long and dicult into one that is just long. 4

The Table Method for Derivatives

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The Table Method for Derivatives