Find the first and second derivatives at the point $(1,2)$, given the function shown below.
[tex]\begin{array}{c}
y^3+2 y-6 x^2+5=11 x^3 \\
y^{\prime}=[?] \\
y^{\prime \prime}=
\end{array}[/tex]
Answer (2)
Use implicit differentiation to find the first derivative y ′ of the given equation.
Evaluate y ′ at the point ( 1 , 2 ) to get y ′ = 14 45 .
Use implicit differentiation again to find the second derivative y ′′ .
Evaluate y ′′ at the point ( 1 , 2 ) to get y ′′ = − 686 2253 .
Explanation
Problem Analysis We are given the equation y 3 + 2 y − 6 x 2 + 5 = 11 x 3 and the point ( 1 , 2 ) . We need to find the first and second derivatives, y ′ and y ′′ , at this point.
First Differentiation First, we differentiate both sides of the equation with respect to x using implicit differentiation. This gives us:
3 y 2 d x d y + 2 d x d y − 12 x = 33 x 2
Solving for y' Now, we solve for d x d y , which we denote as y ′ :
y ′ ( 3 y 2 + 2 ) = 33 x 2 + 12 x
y ′ = 3 y 2 + 2 33 x 2 + 12 x
Evaluating y' at (1,2) Next, we substitute x = 1 and y = 2 into the expression for y ′ to find the value of y ′ at the point ( 1 , 2 ) :
y ′ = 3 ( 2 ) 2 + 2 33 ( 1 ) 2 + 12 ( 1 ) = 12 + 2 33 + 12 = 14 45
Second Differentiation Now, we need to find the second derivative, y ′′ . We differentiate y ′ = 3 y 2 + 2 33 x 2 + 12 x with respect to x using the quotient rule:
y ′′ = ( 3 y 2 + 2 ) 2 ( 66 x + 12 ) ( 3 y 2 + 2 ) − ( 33 x 2 + 12 x ) ( 6 y y ′ )
Evaluating y'' at (1,2) Substitute x = 1 , y = 2 , and y ′ = 14 45 into the expression for y ′′ :
y ′′ = ( 3 ( 2 ) 2 + 2 ) 2 ( 66 ( 1 ) + 12 ) ( 3 ( 2 ) 2 + 2 ) − ( 33 ( 1 ) 2 + 12 ( 1 )) ( 6 ( 2 ) ( 14 45 ))
y ′′ = 1 4 2 ( 78 ) ( 14 ) − ( 45 ) ( 14 540 ) = 196 1092 − 14 24300 = 196 14 15288 − 24300 = 196 14 − 9012 = 14 ⋅ 196 − 9012 = 2744 − 9012 = − 686 2253
Final Answer Therefore, the first derivative at ( 1 , 2 ) is y ′ = 14 45 and the second derivative at ( 1 , 2 ) is y ′′ = − 686 2253 .
Examples
Understanding derivatives is crucial in physics, especially when analyzing motion. For instance, if the given equation described the position of an object over time, the first derivative would represent the object's velocity, and the second derivative would represent its acceleration. By calculating these derivatives at a specific point, we can determine the object's instantaneous velocity and acceleration at that moment, providing valuable insights into its dynamic behavior.
Using implicit differentiation of the given equation, we calculated the first derivative at the point (1,2) as y ′ = 14 45 and the second derivative as ( y'' = -\frac{2253}{686}. \
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