Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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
\[\huge (a+x^{n})^{\frac{ -1 }{ 2 }} \times(n \times x^{n-1}) \]

Answer 11

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

Answer 12

about 2 function?

Answer 13

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)

Answer 14

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

Answer 15

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?

Answer 16

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