find derivative of √3x+3 + 1/√2x^2+4
Answers
d
dxx
3 = lim
∆x→0
(x + ∆x)
3 − x
3
∆x
.
= lim
∆x→0
x
3 + 3x
2∆x + 3x∆x
2 + ∆x
3 − x
3
∆x
.
= lim
∆x→0
3x
2∆x + 3x∆x
2 + ∆x
3
∆x
.
= lim
∆x→0
3x
2 + 3x∆x + ∆x
2 = 3x
2
.
The general case is really not much harder as long as we don’t try to do too much.
The key is understanding what happens when (x + ∆x)
n
is multiplied out:
(x + ∆x)
n = x
n + nxn−1∆x + a2x
n−2∆x
2 + · · · + +an−1x∆x
n−1 + ∆x
n
.
We know that multiplying out will give a large number of terms all of the form x
i∆x
j
, and
in fact that i+j = n in every term. One way to see this is to understand that one method
for multiplying out (x + ∆x)
n
is the following: In every (x + ∆x) factor, pick either the x
or the ∆x, then multiply the n choices together; do this in all possible ways. For example,
for (x + ∆x)
3
, there are eight possible ways to do this:
(x + ∆x)(x + ∆x)(x + ∆x) = xxx + xx∆x + x∆xx + x∆x∆x
+ ∆xxx + ∆xx∆x + ∆x∆xx + ∆x∆x∆x
= x
3 + x
2∆x + x
2∆x + x∆x
2
+ x
2∆x + x∆x
2 + x∆x
2 + ∆x
3
= x
3 + 3x
2∆x + 3x∆x
2 + ∆x
3
No matter what n is, there are n ways to pick ∆x in one factor and x in the remaining
n−1 factors; this means one term is nxn−1∆x. The other coefficients are somewhat harder
to understand, but we don’t really need them, so in the formula above they have simply
been called a2, a3, and so on. We know that every one of these terms contains ∆x to at
least the power 2. Now let’s look at the limit:
d
dxx
n = lim
∆x→0
(x + ∆x)
n − x
n
∆x
= lim
∆x→0
x
n + nxn−1∆x + a2x
n−2∆x
2 + · · · + an−1x∆x
n−1 + ∆x
n − x
n
∆x
= lim
∆x→0
nxn−1∆x + a2x
n−2∆x
2 + · · · + an−1x∆x
n−1 + ∆x
n
∆x
= lim
∆x→0
nxn−1 + a2x
n−2∆x + · · · + an−1x∆x
n−2 + ∆x
n−1 = nxn−1
.
hope it helps you