Question

please help and can you explain step by step for this problem (y+1/y-2)'

Also can anybody tell me if I'm right with the other problem down at the bottom please?

Answer 1

This is the problem? Finding this derivative?\[\Large\bf\sf \left(\frac{y+1}{y-2}\right)'\]

Answer 2

Hold on one moment please.

Answer 3

Thats the problem.

Answer 4

Are both the 1 and x in the numerator?
If so, you need to learn to use brackets to make it clearer XD lol

Answer 5

Yes they are.

Answer 6

\[\Large\bf\sf \left(\frac{1+x}{\sqrt x}\right)'\]We can either quotient rule right away,
or we can separate this into two fractions beforehand.

Answer 7

What method would make more sense to you? :U

Answer 8

Can you do it both ways?

Answer 9

Here is our setup for quotient rule:\[\Large\bf\sf \left(\frac{1+x}{\sqrt x}\right)'\quad=\quad \frac{\color{royalblue}{\left[1+x\right]'}\sqrt x-(1+x)\color{royalblue}{\left[\sqrt x\right]'}}{\left[\sqrt x\right]^2}\]

Answer 10

We have to take the derivative of the blue parts.

Answer 11

Okay.

Answer 12

Do you understand how to calculate the derivative of the first blue part?
For the second part this is an important derivative to remember.\[\Large\bf\sf \frac{d}{dx}\sqrt{x}\quad=\quad \frac{1}{2\sqrt{x}}\]You can write sqrt(x) as x^(1/2) and apply the power rule if you prefer.
But this one shows up so often that it's good to keep in your back pocket.

Answer 13

\[\Large\bf\sf \left(\frac{1+x}{\sqrt x}\right)'\quad=\quad \frac{\color{royalblue}{\left[1+x\right]'}\sqrt x-(1+x)\color{orangered}{\left[\dfrac{1}{2\sqrt x}\right]}}{x}\]

Answer 14

So that's the 1st part?

Answer 15

What? 0_o
That's... the second part.

Answer 16

Okay.

Answer 17

So my answer is in the red? or I have to divide that by x?

Answer 18

I don't understand these questions you're asking...
Have you learned quotient rule? :(

Answer 19

I know it but I forgot it. :/

Answer 20

\[\Large\bf\sf \left(\frac{u}{v}\right)'\quad=\quad \frac{\color{royalblue}{u'}v-u\color{royalblue}{v'}}{v^2}\]

I told you the derivative of the second term,
\[\Large\bf\sf \left(\frac{u}{v}\right)'\quad=\quad \frac{\color{royalblue}{u'}v-u\color{orangered}{v'}}{v^2}\]and plugged it in.

The denominator is x because we had \(\Large\bf\sf (\sqrt x)^2\) which equals x.

Answer 21

So the answer is x?

Answer 22

answer..?

Answer 23

the denominator is x, yes..

Answer 24

Okay.

Answer 25

So that's all or there's more steps?

Answer 26

\[\Large\bf\sf \left(\frac{1+x}{\sqrt x}\right)'\quad=\quad \frac{\color{royalblue}{\left[1+x\right]'}\sqrt x-(1+x)\color{orangered}{\left[\dfrac{1}{2\sqrt x}\right]}}{x}\]You also need the derivative of the first blue part.
Do you know how to differentiate 1+x?

Answer 27

No.

Answer 28

Derivative of 1?

Answer 29

I'm not understanding this.

Answer 30

So the derivative of 1 is -1?

Answer 31

This problem seems to be way beyond your skill level :(
You should start with some easier questions.

No the derivative of 1 is not -1.

Answer 32

The derivative of a constant is zero.

Answer 33

Okay.

Answer 34

So what do I do with the remaining numbers and stuff of the problem?

Answer 35

Maybe the other way would make more sense to you.\[\Large\bf\sf \frac{1+x}{\sqrt x}\quad=\quad \frac{1}{\sqrt x}+\frac{x}{\sqrt x}\]Then rewrite the roots as rational expressions,\[\Large\bf\sf =\frac{1}{x^{1/2}}+\frac{x}{x^{1/2}}\]Rules of exponents allows us to bring the first term up to the numerator by applying a negative to the exponent,\[\Large\bf\sf =x^{-1/2}+\frac{x}{x^{1/2}}\]And for the second term, when we divide things that have similar bases, we `subtract` the exponents.\[\Large\bf\sf =x^{-1/2}+x^{1-1/2}\quad=\quad x^{-1/2}+x^{1/2}\]And from there you can just apply the power rule to each term.
It requires you to remember a lot of your exponent rules though, so maybe it's tougher for you.

Answer 36

Can you explain how to use the power rule?