Find the equation of the tangent to the curve Find the equation of the tangent to the curve $\sqrt X + \sqrt Y = a\;$     at the point $\left(\dfrac {a^2}{4},\dfrac {a^2}{4}\right)$
I don't know how to find $\dfrac {\mathrm dy}{\mathrm dx}$ in this particular question... Please help me to do this sum.
 A: HINT:
$$\sqrt x+\sqrt y=a$$
Differentiating with respect to $x, \frac 1{2\sqrt x}+\frac1{2\sqrt y}\cdot \frac{dy}{dx}=0$ as $\frac{d g(y)}{dx}=\frac{d g(y)}{dy}\cdot\frac{dy}{dx}$
At  $x=y=\frac{a^2}4,$   $$\frac1a+\frac1a\cdot \frac{dy}{dx}=0\implies  \frac{dy}{dx}=-1$$ (assuming $a\ne0$)
A: $$\sqrt x+\sqrt y=a$$
$$ y=(a-\sqrt x)^2$$
$$ y=a^2+x-2a\sqrt x$$
$$\dfrac  {dy}{dx}=0+1-2a\dfrac {1}{2\sqrt x}$$
$$\dfrac  {dy}{dx}=1-\dfrac {a}{\sqrt x}$$
put $x=\dfrac {a^2}{4}$
$$\dfrac  {dy}{dx}_{x=\frac {a^2}{4}}=1-\dfrac {a}{\sqrt{\frac {a^2}{4}}}$$
$$\dfrac  {dy}{dx}=-1$$
A: You do not even need to differentiate, or know anything about derivatives. The curve is symmetrical in $x$ and $y$, and the point we are interested in also does not change when we interchange $x$ and $y$. So the slope must be $1$ or $-1$. A quick sketch rules out the possibility that it is $1$.
A: You can also use this simple equation, although you will have to find the derivative. Obviously by now you will already have your answer but if you can't differentiate an equation the easy way, there is always... First Principles!!
i.e.,
$\frac{f(x+h)-f(x)}{h}$
$$y =f^{\prime}(u)(x-u)+f(u)$$
This gives you the equation of a tangent to curve at a point $(u,y)$
Then plug $\frac{\mathrm dy}{\mathrm dx}$ into this, where $u$ = the $x$ value, expand it and you will have a $y=mx+b$ equation for the tangent!
