How do you find the first and second derivative of $\frac{ln (x^{8})}{x^{2}}$ $ln (x^8)/ x^2$ ?

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Answer 1 · 2017-09-25T11:56:06

$(1) : The First Derivative is, \frac{8}{x^{3}} (x^{2} - 2 ln x) .$ $(1) :" The First Derivative is, "8/x^3(x^2-2lnx).$

$(2) : The second Derivative is, \frac{24}{x^{4}} (2 ln x - x^{2}) .$ $(2) :" The Second Derivative is, "24/x^4(2lnx-x^2).$

Let, $f(x)=ln(x^8)/x^2=(8lnx)/x^2.....[because, ln a^m=mlna],$

$:. f(x)=8*x^-2*lnx.$

By the Product Rule, then, we get,

$f'(x)=8{x^-2d/dx(lnx)+(lnx)d/dx(x^-2)},$

$=8{x^-2*1/x+(-2*x^(-2-1))lnx},$

$=8{x^-1-2x^-3*lnx},$

$=8{1/x-(2lnx)/x^3},$

$rArr f'(x)=8/x^3(x^2-2lnx).$

Knowing that, $f''(x)={f'(x)}',$ we have,

$f''(x)={8(x^-1-2x^-3*lnx)}',$

$=8(x^-1-2x^-3*lnx)',$

$=8(x^-1)'-8*2(x^-3*lnx)',$

$=8(-1*x^(-1-1))-16{x^-3d/dx(lnx)+(lnx)d/dx(x^-3)},$

$=-8x^-2-16{x^-3*1/x+(-3*x^(-3-1))lnx},$

$=-8/x^2-16(1/x^2-(3lnx)/x^4),$

$=-8/x^2-16/x^2+(48lnx)/x^4,$

$=-24/x^2+(48lnx)/x^4.$

$rArr f''(x)=24/x^4(2lnx-x^2).$

Enjoy Maths.!

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