Section C
1) Evaluate the following
i) [tex]$\lim _{x-4}\left(\frac{x^2-16}{x-4}\right)$[/tex]
ii) [tex]$\underset{x-\frac{1}{3}}{\lim }\left(\frac{9 x^2-1}{3 x-1}\right)$[/tex]
b) Differentiate [tex]$(2 x+4)^2(3 x-1)^3$[/tex]
c) If [tex]$x^3+y^3+6 x y=0$[/tex], find [tex]$\frac{d y}{d x}$[/tex] by implicit differentiation
Answer (1)
Evaluate the limit lim x → 4 ( x − 4 x 2 − 16 ) by factoring and direct substitution: 8 .
Evaluate the limit lim x → 3 1 ( 3 x − 1 9 x 2 − 1 ) by factoring and direct substitution: 2 .
Differentiate ( 2 x + 4 ) 2 ( 3 x − 1 ) 3 using the product and chain rules: 2 ( 3 x − 1 ) 2 ( 2 x + 4 ) ( 12 x + 17 ) .
Find d x d y for x 3 + y 3 + 6 x y = 0 using implicit differentiation: y 2 + 2 x − x 2 − 2 y .
Explanation
Problem Overview We are given four problems: two limit evaluations, one differentiation, and one implicit differentiation. We will solve each one step by step.
Evaluating the first limit i) Evaluate lim x → 4 ( x − 4 x 2 − 16 ) .
We can factor the numerator as a difference of squares: x 2 − 16 = ( x − 4 ) ( x + 4 ) .
So, the expression becomes: x − 4 x 2 − 16 = x − 4 ( x − 4 ) ( x + 4 ) .
For x = 4 , we can cancel the ( x − 4 ) terms: x − 4 ( x − 4 ) ( x + 4 ) = x + 4 .
Now, we can evaluate the limit by direct substitution: lim x → 4 ( x + 4 ) = 4 + 4 = 8 .
Evaluating the second limit ii) Evaluate lim x → 3 1 ( 3 x − 1 9 x 2 − 1 ) .
We can factor the numerator as a difference of squares: 9 x 2 − 1 = ( 3 x − 1 ) ( 3 x + 1 ) .
So, the expression becomes: 3 x − 1 9 x 2 − 1 = 3 x − 1 ( 3 x − 1 ) ( 3 x + 1 ) .
For x = 3 1 , we can cancel the ( 3 x − 1 ) terms: 3 x − 1 ( 3 x − 1 ) ( 3 x + 1 ) = 3 x + 1 .
Now, we can evaluate the limit by direct substitution: lim x → 3 1 ( 3 x + 1 ) = 3 ( 3 1 ) + 1 = 1 + 1 = 2 .
Differentiating the function b) Differentiate ( 2 x + 4 ) 2 ( 3 x − 1 ) 3 .
Let f ( x ) = ( 2 x + 4 ) 2 ( 3 x − 1 ) 3 . We will use the product rule: f ′ ( x ) = u ′ ( x ) v ( x ) + u ( x ) v ′ ( x ) , where u ( x ) = ( 2 x + 4 ) 2 and v ( x ) = ( 3 x − 1 ) 3 .
First, find u ′ ( x ) using the chain rule: u ′ ( x ) = 2 ( 2 x + 4 ) ( 2 ) = 4 ( 2 x + 4 ) = 8 x + 16 .
Next, find v ′ ( x ) using the chain rule: v ′ ( x ) = 3 ( 3 x − 1 ) 2 ( 3 ) = 9 ( 3 x − 1 ) 2 .
Now, apply the product rule: f ′ ( x ) = ( 8 x + 16 ) ( 3 x − 1 ) 3 + ( 2 x + 4 ) 2 ( 9 ( 3 x − 1 ) 2 ) .
We can factor out ( 3 x − 1 ) 2 and ( 2 x + 4 ) :
f ′ ( x ) = ( 3 x − 1 ) 2 ( 2 x + 4 ) [ 4 ( 3 x − 1 ) + 9 ( 2 x + 4 )] /2 = ( 3 x − 1 ) 2 ( 2 x + 4 ) [ 6 x − 2 + 18 x + 36 ] = ( 3 x − 1 ) 2 ( 2 x + 4 ) ( 24 x + 34 ) .
f ′ ( x ) = 2 ( 3 x − 1 ) 2 ( 2 x + 4 ) ( 12 x + 17 ) .
Finding dy/dx by implicit differentiation c) If x 3 + y 3 + 6 x y = 0 , find d x d y by implicit differentiation. Differentiate both sides of the equation with respect to x :
d x d ( x 3 ) + d x d ( y 3 ) + d x d ( 6 x y ) = 0 .
3 x 2 + 3 y 2 d x d y + 6 ( x d x d y + y ( 1 ) ) = 0 .
3 x 2 + 3 y 2 d x d y + 6 x d x d y + 6 y = 0 .
Now, solve for d x d y :
( 3 y 2 + 6 x ) d x d y = − 3 x 2 − 6 y .
d x d y = 3 y 2 + 6 x − 3 x 2 − 6 y = y 2 + 2 x − x 2 − 2 y .
Final Answers The solutions are: i) lim x → 4 ( x − 4 x 2 − 16 ) = 8 ii) lim x → 3 1 ( 3 x − 1 9 x 2 − 1 ) = 2 b) d x d (( 2 x + 4 ) 2 ( 3 x − 1 ) 3 ) = 2 ( 3 x − 1 ) 2 ( 2 x + 4 ) ( 12 x + 17 ) c) If x 3 + y 3 + 6 x y = 0 , then d x d y = y 2 + 2 x − x 2 − 2 y
Examples
Limits are used in physics to calculate instantaneous velocity and acceleration. Derivatives are used to find the rate of change of a function, such as the velocity of an object or the slope of a curve. Implicit differentiation is used in related rates problems, such as finding how the rate of change of the radius of a circle is related to the rate of change of its area. These concepts are fundamental in many fields of science and engineering.
