# Given value x, find the tangent line to the curve.

$$y=\frac{2}{3}x^{\frac{3}{2}}+\frac{3}{2}x^{\frac{2}{3}}+1$$ Find the tangent line to the curve at $x=1$. I have taken the derivative of $y$ which was $$y'(x)=\sqrt{x}+\dfrac{1}{\sqrt[3]{x}}$$ I then plugged in the given $x$ value (I think this is the slope of the curve of the tangent line, could someone confirm this?) into $y'(x)$ and got $2$ for the x-coordinate. I then plugged $2$ into the original function $y$ to get the $y$ corrdinate. The problem I have is that my $y$ coordinate does not look correct. I got $$y=\dfrac{4\sqrt{2}}{3}+\dfrac{3\sqrt[3]{4}}{2}+1$$ as the y-coordinate. I know that I haven't finished the problem yet but I am unsure about this step. Is this correct?

-
but what is v?is it variable or constant? – dato datuashvili Feb 27 '13 at 5:42
@dato, looks to me like $v$ is a typo for $x$. – Gerry Myerson Feb 27 '13 at 5:55
Oops, yes the $v$ is supposed to be $x$, sorry. – Kot Feb 27 '13 at 6:03

The derivative evaluated at a particular value will give you the slope of the tangent line at the point which you evaluate it at; in this case, you want to find the tangent line at the point of the graph when $x=1$, so you find $y'(1)=2$. This is the slope of tangent line.

With this information, all we need to do is find a point on this line, and then we can use the point-slope formula for a line and change it to slope-intercept if necessary. But, since we're at $x=1$, we can find that $$y(1)=\frac{2}{3}+\frac{3}{2}+1=\frac{19}{6}.$$ This says the point $(1,19/6)$ is on the line.

So, from the point-slope formula, we have $$y-\frac{19}{6}=2(x-1).$$ Distributing the $2$ and adding $19/6$ over to the other side, we find $$y=2x+\frac{7}{6}.$$

I hope this helps!

-
So the given value was the x-coordinate and not the slope? The derivative of the x-coordinate is the slope which is then used in the point-slope formula. Is that correct? – Kot Feb 27 '13 at 6:05
By the given value, do you mean when the question asks to find the equation of the tangent line at $x=1$? If so, then yes, this is just the $x$-coordinate. We plug this into the derivative to get a slope of the tangent line, then go back and get a point on this line (which we already know has a point in common with the graph of $y$). – Clayton Feb 27 '13 at 6:07
Yes, that was what I meant. Thank you for the explanation! – Kot Feb 27 '13 at 6:16
-edit- I see how you cancelled the exponents out now. :) – Kot Feb 27 '13 at 6:20
@StevenN: I'm glad :) If you have any questions, don't hesitate to ask. – Clayton Feb 27 '13 at 6:26