# 009A Sample Midterm 1, Problem 4 Detailed Solution

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Let  ${\displaystyle y={\sqrt {3x-5}}.}$

(a) Use the definition of the derivative to compute   ${\displaystyle {\frac {dy}{dx}}}$   for  ${\displaystyle y={\sqrt {3x-5}}.}$

(b) Find the equation of the tangent line to  ${\displaystyle y={\sqrt {3x-5}}}$  at  ${\displaystyle (2,1).}$

Background Information:
1. Recall
${\displaystyle f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}$
2. The equation of the tangent line to  ${\displaystyle f(x)}$  at the point  ${\displaystyle (a,b)}$  is
${\displaystyle y=m(x-a)+b}$  where  ${\displaystyle m=f'(a).}$

Solution:

(a)

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

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3(x+h)-5}}-{\sqrt {3x-5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3x+3h-5}}-{\sqrt {3x-5}}}{h}}.}\end{array}}}$

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

(b)

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

(a)     ${\displaystyle {\frac {dy}{dx}}={\frac {3}{2{\sqrt {3x-5}}}}}$
(b)     ${\displaystyle y={\frac {3}{2}}(x-2)+1}$