Question

derivative of x^2 ln(sqrt(2x+1)

Answer 1

you should use the product rule d(u v) = u dv + v du
and the chain rule

Answer 2

i have the first half of my equation right but ii'm doubting the second part. this was my final equation=
2x ln(sqrt(2x+1))+ x^2. (4x^2 + 4x + 1)^-1.5

Answer 3

yes the first term is ok
For the 2nd term , we will use d ln(u) = 1/u du
the 2nd term is
\[ x^2 \frac{d}{dx} \ln(u) = x^2 \frac{1}{u} \ \frac{d}{dx} u \]
where \( u = \sqrt{2x+1} =\left(2x+1\right)^\frac{1}{2} \)

we get
\[ \frac{x^2}{\left(2x+1\right)^\frac{1}{2}} \frac{d}{dx}\left(2x+1\right)^\frac{1}{2} \]
can you finish ?

Answer 4

i'm seriously lost, the notation kind of confuses me =/

Answer 5

d/dx = 0.5 (2x+1)^-1.5

Answer 6

so i'm thinking x^2/(2x+1)^0.5 . 0.5 (2x+1)^-1.5 . 2

Answer 7

the derivative of u^n = n u^(n-1)
\[ d u^n = n \ u^{n-1} du \]
that says: to take the derivative of u to the nth power, multiply u by n, and decrement the exponent by 1. then multiply all of that by the derivative of u (with respect to whatever variable we are using... in this case with respect to x)

Answer 8

*** d/dx = 0.5 (2x+1)^-1.5 ***

How are you getting an exponent of -1.5 ?
you should be doing (n-1) where n is 1/ 2

Answer 9

ahh good lord, so it should be 0.5 (2x+1)^-1/2
so x^2/(2x+1)^0.5 . 0.5 (2x+1)^-1/2 . 2

Answer 10

which can be simplified

Answer 11

\[ \frac{x^2}{\sqrt{2x+1} } \cdot \frac{1}{\cancel2} \frac{1}{\sqrt{2x+1} } \cdot \cancel2 \\
=\frac{x^2}{2x+1}
\]
that is the 2nd term. The full answer is
\[ 2x\ \ln\left( \sqrt{2x+1}\right) + \frac{x^2}{2x+1} \]

Answer 12

sry, just 1 last question, the (x^2/ sqrt(2x+1)) . (1/ sqrt(2x+1)), how does that turn in to just (x^2/ sqrt(2x+1))

Answer 13

when you multiply fractions, you multiply top times top and bottom times bottom

for the bottom we use
\[ \sqrt{x} \cdot \sqrt{x} = \sqrt{x \cdot x} = \sqrt{x^2} = x \]

Answer 14

the bottom is not sqr(2x+1) it is just (2x+1)

Answer 15

oh now i see it, THANK YOU VERY MUCH!!

Answer 16

yw