Question

y = 10th sqrt of (5x^2 + 3)/(2x^4 - 6) ? how do i find the derivative using logarithmic diffrientation?

Answer 1

I suppose that you mean \[\sqrt[10]{(5x^2+3)/(2x^4-6)}\]

Answer 2

Simplify using the quotient rule with the chain rule. Turn the radical into an exponent.

Answer 3

yes sir..

Answer 4

you take the log of both sides and then use implicit differentiation I think cus remember that logx^x is the same as xlogx understand?

Answer 5

ln(y)=(1/10)ln((5x^2 + 3)/(2x^4 - 6)) then you take the derivative of y with respect to t and x with respect to t so (1/y)*y'= (1/10)(1/((5x^2 + 3)/(2x^4 - 6)))((10x*x')*(2x^4 - 6)^-1+(5x^2 + 3)*(-1)(2x^4 - 6)^-2(8x^3*x')) okay I didn't see the division at first this is a pretty messy problem hopefully I am doing this right y'= y*(1/10)(1/((5x^2 + 3)/(2x^4 - 6)))(x'((10x)*(2x^4 - 6)^-1+(5x^2 + 3)*(-1)(2x^4 - 6)^-2(8x^3))) then I think you just put in the original equation for y but I may have messed this up I am pretty new to implicit differentiation does this like okay to everyone? I feel like that pesky x' shouldn't be there.