$$y=\frac{2}{3}x^{\frac{3}{2}}+\frac{3}{2}x^{\frac{2}{3}}+1$$
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?
[Math] Given value x, find the tangent line to the curve.
calculusderivatives
Best Answer
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!