2
$\begingroup$

I am suppose to find the equation of a tangent line and I am given the following information: $ y = \sqrt[4] {x}$, $(1,1)$ I know the formula $f(a+h)-f(a)$ but it does not seem to be helping me.

  • 0
    So you have already learned the rules for computing derivatives, don't you? In your question [63684](http://math.stackexchange.com/q/63684/752) you had not. As Srivatsan Narayanan commented you are right.2011-09-18

2 Answers 2

2

Equation of tangent line at point $(a,f(a))$ is y = f(a) + f'(a)(x - a), so we have to find f'(x) and than plug in value $a$ into the result.

f'(x) = (\sqrt[4]{x})' = ({x^{\frac{1}{4}}})' = \frac{1}{4}{x^{\frac{{ - 3}}{4}}}

\implies f'(a) = f'(1) = \frac{1}{4}

Since $f(1) = 1$, we can write next equation $y = 1 + \frac{1}{4}(x - 1)$

which means that equation of tangent line is $y = \frac{1}{4}x + \frac{3}{4}$.

  • 0
    @JordanCarlyon,with this kind of given informations I don't know the other way to find a tangent line2011-09-26
1

SET OF HINTS:

For the point-slope form of a tangent lint, $y - y_0 = m (x - x_0)$, you need exactly 2 pieces of information: the slope $m$ of the line and a point $(x_0, y_0)$ that the line goes through.

You have the point, so how do you find the slope? Well, the derivative tells you something about the slope, right? If you know what that is, then you can assemble both of these pieces of information into the equation for the tangent line.

  • 1
    I have the correct answer in the back o$f$ my book, but not the ability to get to it on my own yet.2011-09-18