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

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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
i still did not get it.. can u teach me in detail?
I don't know I will be able to complete, thre is a chance of power failure here.. but let me try

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ok... as a first we need to know chain rule of differentiation
as an example \[\huge \frac{ d }{ dx } (\sqrt{a+x^{n}}) = \frac{ d }{ dx } ({a+x^{n}})^\frac{ 1 }{ 2 }\]
a function of x... and find the derivative by function by function using the chain rule...(function by function differntiation)
here first function is square root (raised to 1/2) second function of x is \[\large x^{n}\] chain rule states that \[\huge \frac{ d }{ dx }(f(g(x))) = f^{'}(g(x))\times g^{'}(x)\]
here in this example square root is f(x) and \[\large x^{n}\] is g(x) Hope that now you can some what understand
i think i get a little what do u say...
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 \[\huge (a+x^{n})^{\frac{ -1 }{ 2 }} \times(n \times x^{n-1}) \]
hope you under stood... so comming back to your question can you say how many functions are there in your qustion?
about 2 function?
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)
yes you are right... two functions... I think powerfailure will be happening in a minute :(
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?
yes .. it is right......(sorry for late reply) glad you figured out

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