009A Sample Midterm 1, Problem 4 Detailed Solution

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Let  $y={\sqrt {3x-5}}.$ (a) Use the definition of the derivative to compute   ${\frac {dy}{dx}}$ for  $y={\sqrt {3x-5}}.$ (b) Find the equation of the tangent line to  $y={\sqrt {3x-5}}$ at  $(2,1).$ Background Information:
1. Recall
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}$ 2. The equation of the tangent line to  $f(x)$ at the point  $(a,b)$ is
$y=m(x-a)+b$ where  $m=f'(a).$ Solution:

(a)

Step 1:
Let  $f(x)={\sqrt {3x-5}}.$ Using the limit definition of the derivative, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3(x+h)-5}}-{\sqrt {3x-5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3x+3h-5}}-{\sqrt {3x-5}}}{h}}.}\end{array}}$ Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\sqrt {3x+3h-5}}-{\sqrt {3x-5}})}{h}}{\frac {({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}{({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {(3x+3h-5)-(3x-5)}{h({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3h}{h({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3}{{\sqrt {3x+3h-5}}+{\sqrt {3x-5}}}}}\\&&\\&=&\displaystyle {\frac {3}{{\sqrt {3x-5}}+{\sqrt {3x-5}}}}\\&&\\&=&\displaystyle {{\frac {3}{2{\sqrt {3x-5}}}}.}\end{array}}$ (b)

Step 1:
We start by finding the slope of the tangent line to  $f(x)={\sqrt {3x-5}}$ at  $(2,1).$ Using the derivative calculated in part (a), the slope is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {\frac {3}{2{\sqrt {3(2)-5}}}}\\&&\\&=&\displaystyle {{\frac {3}{2}}.}\end{array}}$ Step 2:
Now, the tangent line to  $f(x)={\sqrt {3x-5}}$ at  $(2,1)$ has slope  $m={\frac {3}{2}}$ and passes through the point  $(2,1).$ Hence, the equation of this line is
$y={\frac {3}{2}}(x-2)+1.$ (a)     ${\frac {dy}{dx}}={\frac {3}{2{\sqrt {3x-5}}}}$ (b)     $y={\frac {3}{2}}(x-2)+1$ 