Question

find the derivative of f(x)=x^2 ln x

Answer 1

\[(fg)'-f'g+g'f\] with
\[f(x)=x^2,f'(x)=2x,g(x)=\ln(x), g'(x)=\frac{1}{x}\]

Answer 2

.y = (lnx)/x^2
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y = f(x)/g(x)
dy/dx = (f' *g - g' *f)/g^2
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dy/dx = [(1/x) *x^2 - (2x) lnx] / x^4
dy/dx = [x - (2x) lnx] / x^4
dy/dx = [1 - 2 lnx] / x^3 >>>>
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d^y/dx^2 = [-(2/x)*x^3 - 3x^2(1 - 2 lnx)] / x^6
d^y/dx^2 = - [2 + 3(1 - 2 lnx)] / x^4
d^y/dx^2 = - [5 -6 lnx)] / x^4
d^y/dx^2 = [- 5 +6 lnx] / x^4 >>>

Answer 3

so f'(x)=x+2x lnx

Answer 4

yup

Answer 5

Thanks

Answer 6

wy