Finding tanget line to a graph at given point

**eniav** · March 1st 2008, 03:54 PM

Find the equation of the tangent line to the graph of the given function at the given value of x,
F(x)=((x^2)+28)^(4/5), x=2

**bobak** · March 1st 2008, 03:59 PM

$f(x) = (x^2 + 28)^{\frac{4}{5}}$

$f'(x) = \frac{4}{5} 2x (x^2 + 28)^{- \frac{1}{5}}$ by the chain rule

$f'(2) = \frac{16}{5} (32)^{- \frac{1}{5}}$
$f'(2) = \frac{8}{5}$

the rest is pretty routine

$y - f(2) = f'(2) ( x - 2)$

can you finish that off ?

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