need help with Find the equation of the tangent line

[warsatan]

I have this problem, trying to work it out, can someone point out what i do wrong? thanks

Find the equation of the tangent line, in slope-intercept form, to the graph of
f(x) = (1/3)x squareroot of x²+5 at the point (2,2).

I rewrite f(x) = (1/3)x (x² + 5)^1/2

this is what i got so far,

f'(x)= (1/3)x [(x²+5)^1/2 ] + (x²+5)^1/2 (1/3)

= (1/3)x [(1/2)(x² +5)^(-1/2)(2x)] +(1/3)(x²+5)^1/2

= (1/6)2x²(x²+5)^-1/2 + (1/3) (x²+5)^1/2

= x²(x²+5)^(-1/2) + (x²+5)^1/2
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3


What mistake did i make? can some one help me point it out? thanks alot.

 

[stapel]

How does "(1/3) x (x² + 5)^1/2" equal "(1/3) x (x² + 5)^1/2 + (1/3) (x² + 5)^1/2"? This would require that (1/3) (x² + 5)^1/2 equal zero, and I see nothing in the exercise that justifies this assumption...?

Eliz.

 

[warsatan]

sorry about that mislead, it meant to be derivative of f(x)