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sha0403
find dy/dx for sqrt of (9+ ((x-9)/6)^1/5) ? i don't know how to do this...
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
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?
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