Question

Find the first and second
derivatives at the point (1, 3),
given the function shown below.
y² + 3 + 2x = 14x³
20
y' = 20/3
y" = [?]
Round your answer to the nearest thousandth.

Answer 1

Answer:

$y'=\dfrac{20}{3}=6.667\; \sf (nearest\;thousandth)$

$y''=\dfrac{126}{29}=4.345\; \sf (nearest\;thousandth)$

Step-by-step explanation:

Given function:

$y^2+3+2x=14x^3$

To differentiate an equation that contains a mixture of x and y terms, we can use implicit differentiation.

Begin by placing d/dx in front of each term of the equation:

$\dfrac{\text{d}}{\text{d}x}y^2+\dfrac{\text{d}}{\text{d}x}3+\dfrac{\text{d}}{\text{d}x}2x=\dfrac{\text{d}}{\text{d}x}14x^3$

Differentiate the terms in x only (and the constant term):

$\dfrac{\text{d}}{\text{d}x}y^2+0+2=14\cdot 3x^{3-1}$

$\dfrac{\text{d}}{\text{d}x}y^2+2=42x^2$

Use the chain rule to differentiate terms in y only.

In practice, this means differentiate with respect to y, and place dy/dx at the end:

$2y\dfrac{\text{d}y}{\text{d}x}+2=42x^2$

Rearrange the resulting equation to make dy/dx the subject:

$2y\dfrac{\text{d}y}{\text{d}x}=42x^2-2$

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{42x^2-2}{2y}$

Simplify by dividing the numerator and denominator by 2:

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{21x^2-1}{y}$

Therefore, the first derivative is:

$\large\boxed{\boxed{y'=\dfrac{21x^2-1}{y}}}$

Now, substitute x = 1 and y = 3 to find the first derivative at point (1, 3):

$y'=\dfrac{21(1)^2-1}{3}$

$y'=\dfrac{20}{3}$

Rounding to the nearest thousandth gives:

$y'=6.667$

$\hrulefill$

To find the second derivative, we need to differentiate the first derivative using the quotient rule:

$\boxed{\begin{array}{c}\underline{\textsf{Quotient Rule for Differentiation}}\\\\\textsf{If y=\dfrac{u}{v} then:}\\\\\dfrac{\text{d}y}{\text{d}x}=\dfrac{v \dfrac{\text{d}u}{\text{d}x}-u\dfrac{\text{d}v}{\text{d}x}}{v^2}\\\\\end{array}}$

In this case:

$u = 21x^2 - 1$

$v = y$

Differentiate u and v separately:

$\dfrac{\text{d}u}{\text{d}x} = 2\cdot21x^{2-1}-0=42x$

$\dfrac{\text{d}v}{\text{d}x} = 1 \cdot \dfrac{\text{d}y}{\text{d}x} = \dfrac{\text{d}y}{\text{d}x}$

Now, put everything into the quotient rule formula:

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{y \cdot 42x-(21x^2-1) \cdot \dfrac{\text{d}y}{\text{d}x}}{y^2}$

Simplify:

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{42xy-(21x^2-1) \dfrac{\text{d}y}{\text{d}x}}{y^2}$

Rearrange to isolate dy/dx:

$y^2\dfrac{\text{d}y}{\text{d}x}=42xy-(21x^2-1) \dfrac{\text{d}y}{\text{d}x}$

$y^2\dfrac{\text{d}y}{\text{d}x}+(21x^2-1) \dfrac{\text{d}y}{\text{d}x}=42xy$

$\dfrac{\text{d}y}{\text{d}x}(y^2+(21x^2-1))=42xy$

$\dfrac{\text{d}y}{\text{d}x}(21x^2+y^2-1)=42xy$

$\dfrac{\text{d}y}{\text{d}x}=\dfrac{42xy}{21x^2+y^2-1}$

Therefore, the second derivative is:

$\large\boxed{\boxed{y''=\dfrac{42xy}{21x^2+y^2-1}}}$

Now, substitute x = 1 and y = 3 to find the second derivative at point (1, 3):

$y''=\dfrac{42(1)(3)}{21(1)^2+(3)^2-1}$

$y''=\dfrac{126}{21+9-1}$

$y''=\dfrac{126}{29}$

Rounding to the nearest thousandth gives:

$y''=4.345$