Question

differentiate the function. then find an equation of the tangent line at the indicated point on the graph of the function. w = g(z)=7 + \sqrt{16 - z}, (z,w)=(7,10)
the derivative of the function w = g(z)=7 + \sqrt{16 - z} is . (type an exact answer, using radicals as needed.)

Explanation:

Step1: Rewrite the function

Let $y = 7+(16 - z)^{\frac{1}{2}}$.

Step2: Apply the sum - rule of differentiation

The derivative of a sum $y = u + v$ is $y^\prime=u^\prime + v^\prime$. Here $u = 7$ and $v=(16 - z)^{\frac{1}{2}}$. The derivative of a constant $u = 7$ is $u^\prime=0$.

Step3: Apply the chain - rule to differentiate $v=(16 - z)^{\frac{1}{2}}$

Let $u = 16 - z$, so $v = u^{\frac{1}{2}}$. First, $\frac{dv}{du}=\frac{1}{2}u^{-\frac{1}{2}}$ and $\frac{du}{dz}=- 1$. By the chain - rule $\frac{dv}{dz}=\frac{dv}{du}\cdot\frac{du}{dz}$. Substituting $u = 16 - z$ back in, we get $\frac{dv}{dz}=\frac{1}{2}(16 - z)^{-\frac{1}{2}}\cdot(-1)=-\frac{1}{2\sqrt{16 - z}}$.

Step4: Find the derivative of the whole function

Since $y^\prime=u^\prime + v^\prime$ and $u^\prime = 0,v^\prime=-\frac{1}{2\sqrt{16 - z}}$, then $g^\prime(z)=-\frac{1}{2\sqrt{16 - z}}$.

Answer:

$-\frac{1}{2\sqrt{16 - z}}$