#f(5)=frac{5sqrt(30)}{12}#
#f'(x)=frac{d}{dx}[x^2(x^3-5)^{-1/2}]#
#=x^2frac{d}{dx}[(x^3-5)^{-1/2}]+(x^3-5)^{-1/2}frac{d}{dx}(x^2)#
#=x^2[(-1/2)(x^3-5)^{-3/2}(3x^2)]+(x^3-5)^{-1/2}(2x)#
#=frac{x(x^3-20)}{2(x^3-5)^{3/2}}#
#f'(5)=frac{7sqrt(30)}{192}#
Let the equation of the tangent line be
#y(x)=mx+c#,
where #m# and #c# are constants to be determined.
For a line to be tangent to a curve, they must intercept at the point of interest.
#y(5)=f(5)#
#5m+c=frac{5sqrt(30)}{12}#
In addition, the line and the curve should have the same gradient at the point of interception.
#y'(5)=f'(5)#
#m=frac{7sqrt(30)}{192}#
Solving for #c#,
#c=frac{5sqrt(30)}{12}-5frac{7sqrt(30)}{192}#
#=frac{15sqrt(30)}{64}#
#y(x)=frac{7sqrt(30)}{192}x+frac{15sqrt(30)}{64}#
