**Theorem 1:** Suppose *g* is differentiable on an open interval containing *x=c*. If both and exist, then the two limits are equal, and the common value is *g'(c)*.

**Proof:** Let and . By the Mean Value Theorem, for every positive *h* sufficiently small, there exists satisfying such that: .

Then: .

Similarly, for every positive *h* sufficiently small, there exists satisfying such that: .

Then: . This shows that .

Note: The same proof can be modified to show that if *g* is continuous at *x=c* and differentiable on both sides of *x=c*, and if , then *g* is differentiable at *x=c* with *g'(c)=L*.

**Theorem 2:** Suppose *p* and *q* are defined on an open interval containing *x=c*, and each are differentiable at *x=c*.

Let:

Then *f* is differentiable at *x=c* if and only if *p(c)=q(c)* and *p'(c)=q'(c)*.

**Proof:** We know that *f'(c)* exists if and only if .

We have that: .

Also: if and only if *p(c)=q(c)*. So *f* will be differentiable at *x=c* if and only if *p(c)=q(c)* and *p'(c)=q'(c)*.

## 2003 AB6, part (c)

Suppose the function *g* is defined by: where *k* and *m* are constants. If *g* is differentiable at *x=3* what are the values of *k* and *m*?

### Method 1:

We are told that *g* is differentiable at *x=3*, and so *g* is certainly differentiable on the open interval (0,5).

and .

So the two limits both exist and by Theorem 1 must be equal. Hence . Since *g* is continuous at *x=3, 2k=3m+2*. This gives the two equations to solve for *k* and *m*.

### Method 2:

Let and *q(x)=mx+2*. Both are differentiable at *x=3*. If *g* is differentiable at *x=3*, then Theorem 2 implies that *p(3)=q(3)* and *p'(3)=q'(3)*. This yields the two same two equations as Method 1.

Either the note after Theorem 1 or Theorem 2 can be used to show that if we *choose* and , then we can prove that *g* is differentiable at *x=3*.

*Jim Hartman teaches at the College of Wooster in Wooster, Ohio.*

*Larry Riddle is a professor of mathematics at Agnes Scott College in Decatur, Georgia.*