Question

Find derivative
y=1/square root of x^2-7x+3

Answer 1

Are you still here?

Answer 2

yes

Answer 3

Are you pretty familiar with negative exponents, techdodo?

Answer 4

yes

Answer 5

Fractional exponents, too?

Answer 6

yea

Answer 7

\[\frac{1}{\sqrt{x^2-7 x+3}}\]
\[\left(x^2-7 x+3\right)^{\frac{-1}{2}}\]
You will have to use chain rule
first take derivative of outer function, don't mind what inside.
\[\frac{-1}{2}\left(\text{ }x^2-7 x+3\text{ }\right)^{\frac{-3}{2}}\]

Now take inner derivative
2x-7

Multiply inner and outer derivative
\[\frac{-1}{2}\left(\text{ }x^2-7 x+3\text{ }\right)^{\frac{-3}{2}}(2x-7)\]

Answer 8

Excellent! Let's do this one step at a time, then:
y=1/square root of x^2-7x+3
Can you show me what this would look like in pure exponential form?

Answer 9

Still here, dodo?

Answer 10

Ah, oh well. Imran showed you the exact steps I was going to walk you through, so just follow those carefully.

Answer 11

yea was just having trouble with the radical

Answer 12

but i forgot bout bringing it up and making the exponent negative

Answer 13

O ok. Well let's do it one step at a time. How do you transform the radical into an exponent?

Answer 14

thank you though

Answer 15

O ok. Glad to know you knew it. :D

Answer 16

(x^2-7x+3)^1/2

Answer 17

Yup :D And then you just do the negative, and you're all set to chain rule. Great work. :D

Answer 18

just the exponent would be negative when it moves up right?

Answer 19

Correct. Because 1/(anything) means (anything)^-1

Answer 20

Likewise, 1/((anything)^2) would equal (anything)^-2