Tangent Lines and Curves Write an equation for the line tangent to the graph of x=y^2+4 at the point (5,1).
I have found the derivative y'=1/2y but I do not know what to do next.
 A: Find $y'$ at the point $(5, 1)$. That is the slope of the line. Then you can use point-slope formula.
A: Hint: In general, the tangent to the curve $y=f(x)$ at $(x_1,y_1)$ is
$$y=y_1+\left.\dfrac{dy}{dx}\right|_{x=x_1}(x-x_1)=y_1+f^\prime(x_1)(x-x_1)$$
A: Since $y^{\prime}=\frac{1}{2y}$, find $y^{\prime}$ by substitute (5, 1) into the equation which is $y^{\prime} = \frac{1}{2}$ = gradient of the equation.
Equation of a line: $$\begin{align}y-y_1=m(x-x_1)\end{align}$$
Thus, the equation should be $$\begin{align}y-1=\frac{1}{2}(x-5)\end{align}$$
Tidy it up and becomes $$\begin{align}x-2y-3=0\end{align}$$
A: The curve is $x=y^2+4$, it indeed contains $(5,1)$. Its differential is
$$dx=2y\,dy$$
(Consequently, $\displaystyle\frac{dy}{dx}\ =\ \frac1{2y}\ $ indeed.)
Now, at $y=1$, it gives $dy/dx=1/2$, this is the slope of the curve at that point (note that value of $y$ uniquely determines value of $x$).
Finally, the equation of the tangent will be of the form
$$y=\frac12\,x+c$$
and it goes through $(5,1)$, so $x=5$ and $y=1$ should satisfy it, this gives you the constant part $c$:
$$1=\frac52\,+c\,.$$
