Question

differentiate the function. then find an equation of the tangent line at the indicated point on the graph of the function. y = f(x)=5 + \sqrt{3 - x}, (x,y)=(-6,8)
the derivative of the function y = f(x)=5 + \sqrt{3 - x} is

Explanation:

Step1: Rewrite the function

Rewrite $y = 5+\sqrt{3 - x}=5+(3 - x)^{\frac{1}{2}}$.

Step2: Apply the sum - rule of differentiation

The derivative of a sum $y = u + v$ is $y'=u'+v'$. Here $u = 5$ and $v=(3 - x)^{\frac{1}{2}}$. The derivative of a constant $u = 5$ is $u'=0$.

Step3: Apply the chain - rule to differentiate $v=(3 - x)^{\frac{1}{2}}$

Let $u = 3 - x$, then $v = u^{\frac{1}{2}}$. First, find $\frac{dv}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dx}=-1$. By the chain - rule $\frac{dv}{dx}=\frac{dv}{du}\cdot\frac{du}{dx}$. Substituting $u = 3 - x$ back in, we have $\frac{dv}{dx}=\frac{1}{2}(3 - x)^{-\frac{1}{2}}\cdot(-1)=-\frac{1}{2\sqrt{3 - x}}$.

Step4: Find the derivative of $y$

Since $y'=u'+v'$ and $u' = 0$, $v'=-\frac{1}{2\sqrt{3 - x}}$, then $y'=-\frac{1}{2\sqrt{3 - x}}$.

Answer:

$-\frac{1}{2\sqrt{3 - x}}$