Question

Find the nth derivative of x cube log x

Answer 1

Answer:

here is the answer

Step-by-step explanation:

y= x^3 * logx ….Assume logx by definition = ln x / ln 10

y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)

= 3x^2 log x +x^2 (1/ ln 10)

y” = 6x log x + 3x^2 (1/ln 10)*(1/x) +2x / ( ln 10)

= 6x log x + 3x(1/ln 10)*(1/x) +2x / ( ln 10)

y”’ = 6 log x+6/ln(10) + 5/ (ln 10) =

y”” =( 6/ ln 10)* (1/x)

Answer 2

y= x^3 * logx ….Assume logx by definition = ln x / ln 10

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)= 3x^2 log x +x^2 (1/ ln 10)

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)= 3x^2 log x +x^2 (1/ ln 10)y” = 6x log x + 3x^2 (1/ln 10)*(1/x) +2x / ( ln 10)

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)= 3x^2 log x +x^2 (1/ ln 10)y” = 6x log x + 3x^2 (1/ln 10)*(1/x) +2x / ( ln 10)= 6x log x + 3x(1/ln 10)*(1/x) +2x / ( ln 10)

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)= 3x^2 log x +x^2 (1/ ln 10)y” = 6x log x + 3x^2 (1/ln 10)*(1/x) +2x / ( ln 10)= 6x log x + 3x(1/ln 10)*(1/x) +2x / ( ln 10)y”’ = 6 log x+6/ln(10) + 5/ (ln 10) =

y= x^3 * logx ….Assume logx by definition = ln x / ln 10y' = 3x^2 log x +x^3 (1/ ln 10)× (1/x)= 3x^2 log x +x^2 (1/ ln 10)y” = 6x log x + 3x^2 (1/ln 10)*(1/x) +2x / ( ln 10)= 6x log x + 3x(1/ln 10)*(1/x) +2x / ( ln 10)y”’ = 6 log x+6/ln(10) + 5/ (ln 10) =y”” =( 6/ ln 10)* (1/x)!-