## anonymous 3 years ago find dy/dx for sqrt of (9+ ((x-9)/6)^1/5) ? i don't know how to do this...

1. anonymous

I think nesting of differentiation is a good idea.. $\huge \frac{ d }{ dx } (\sqrt{9+(\frac{ x-9 }{ 6 })^{\frac{ 1 }{ 5 }}})$ $\huge \frac{ d }{ dx } (x^{n+1}) = (n+1)x^{n}$ then use the chain rule $\huge \sqrt{x} = x^{\frac{ 1 }{ 2}}$ hope this will assist you

2. anonymous

i still did not get it.. can u teach me in detail?

3. anonymous

I don't know I will be able to complete, thre is a chance of power failure here.. but let me try

4. anonymous

ok... as a first we need to know chain rule of differentiation

5. anonymous

as an example $\huge \frac{ d }{ dx } (\sqrt{a+x^{n}}) = \frac{ d }{ dx } ({a+x^{n}})^\frac{ 1 }{ 2 }$

6. anonymous

a function of x... and find the derivative by function by function using the chain rule...(function by function differntiation)

7. anonymous

here first function is square root (raised to 1/2) second function of x is $\large x^{n}$ chain rule states that $\huge \frac{ d }{ dx }(f(g(x))) = f^{'}(g(x))\times g^{'}(x)$

8. anonymous

here in this example square root is f(x) and $\large x^{n}$ is g(x) Hope that now you can some what understand

9. anonymous

i think i get a little what do u say...

10. anonymous

so the answer for the example becomes $\huge (a+x^{n})^{\frac{ -1 }{ 2 }} \times (0+n \times x^{n-1})$ which is equal to $\huge (a+x^{n})^{\frac{ -1 }{ 2 }} \times(n \times x^{n-1})$

11. anonymous

hope you under stood... so comming back to your question can you say how many functions are there in your qustion?

12. anonymous

13. anonymous

try to figure out what is the functions... and how you can apply chain rule.... and refer example when in doubt... i listd the needed formule in my first reply...(expect chain rule, which is listed in above comments)

14. anonymous

yes you are right... two functions... I think powerfailure will be happening in a minute :(

15. anonymous

ok..then i try i get 1/2((9+(x-9)/6)^1/5)^-1/2 . 1/30((x-9)/6)^-4/5) its right or not?

16. anonymous

yes .. it is right......(sorry for late reply) glad you figured out