Question

1. Find dy/dx when y= 6x²y³ + 2x²y² at (1) 2. Find dy/dx when y= x² +y² - 25 Please Explain In Depth.

asked Jun 13, 2024 64.5k views

1 Answer

Roomtek · Answer 1 · 2024-06-17T04:34:48+0000

Answer:

$\textsf{1)} \quad \frac{\text{d}y}{\text{d}x}=(√(7))/(7)\;\;\;\textsf{and}\;\;\;\frac{\text{d}y}{\text{d}x}=-(√(7))/(7)$

$\textsf{2)} \quad \frac{\text{d}y}{\text{d}x}=(2x)/(1-2y)$

Explanation:

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

Question 1

To differentiate y = 6x²y³ + 2x²y², begin by placing d/dx in front of each term:

$\frac{\text{d}}{\text{d}x}y=\frac{\text{d}}{\text{d}x}6x^2y^3+\frac{\text{d}}{\text{d}x}2x^2y^2$

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:

$\frac{\text{d}y}{\text{d}x}=\frac{\text{d}}{\text{d}x}6x^2y^3+\frac{\text{d}}{\text{d}x}2x^2y^2$

Use the product rule to differentiate the terms in x and y.

$\boxed{\begin{minipage}{5.5 cm}\underline{Product Rule for Differentiation}\\\\If $y=uv$ then:\\\\$\frac{\text{d}y}{\text{d}x}=u\frac{\text{d}v}{\text{d}x}+v\frac{\text{d}u}{\text{d}x}$\\\end{minipage}}$

$\textsf{To differentiate}\;6x^2y^3:$

$\textsf{Let}\;u=6x^2\implies\frac{\text{d}u}{\text{d}x}=12x$

$\textsf{Let}\;v=y^3\implies\frac{\text{d}v}{\text{d}x}=3y^2\frac{\text{d}y}{\text{d}x}$

Therefore:

$\begin{aligned}\frac{\text{d}}{\text{d}x}6x^2y^3&=6x^2\cdot3y^2\frac{\text{d}y}{\text{d}x}+y^3\cdot12x\\\\&=18x^2y^2\frac{\text{d}y}{\text{d}x}+12xy^3\end{aligned}$

$\textsf{To differentiate}\;2x^2y^2:$

$\textsf{Let}\;u=2x^2\implies\frac{\text{d}u}{\text{d}x}=4x$

$\textsf{Let}\;v=y^2\implies\frac{\text{d}v}{\text{d}x}=2y\frac{\text{d}y}{\text{d}x}$

Therefore:

$\begin{aligned}\frac{\text{d}}{\text{d}x}2x^2y^2&=2x^2\cdot2y\frac{\text{d}y}{\text{d}x}+y^2\cdot4x\\\\&=4x^2y\frac{\text{d}y}{\text{d}x}+4xy^2\end{aligned}$

So the final differentiated equation is:

$\frac{\text{d}y}{\text{d}x}=18x^2y^2\frac{\text{d}y}{\text{d}x}+12xy^3+4x^2y\frac{\text{d}y}{\text{d}x}+4xy^2$

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

$\frac{\text{d}y}{\text{d}x}-18x^2y^2\frac{\text{d}y}{\text{d}x}-4x^2y\frac{\text{d}y}{\text{d}x}=12xy^3+4xy^2$

$\frac{\text{d}y}{\text{d}x}(1-18x^2y^2-4x^2y)=12xy^3+4xy^2$

$\frac{\text{d}y}{\text{d}x}=(12xy^3+4xy^2)/(1-18x^2y^2-4x^2y)$

To find dy/dx when x = 1, first find the value of y when x = 1 by substituting x = 1 into the original equation:

$\begin{aligned}x=1\implies y&=6(1)^2y^3+2(1)^2y^2\\y&=6y^3+2y^2\\0&=6y^3+2y^2-y\\0&=y(6y^2+2y-1)\end{aligned}$

Therefore:

y=0

6y^2+2y-1=0

Use the quadratic formula to solve the quadratic:

$y=(-2\pm√(2^2-4(6)(-1)))/(2(6))$

$y=(-2\pm2√(7))/(12)$

$y=(-1\pm√(7))/(6)$

Therefore, the points on the curve y = 6x²y³ + 2x²y² when x = 1 are:

$(1,0)\;\;\textsf{and}\;\;\left(1,(-1-√(7))/(6)\right)\;\;\textsf{and}\;\;\left(1,(-1+√(7))/(6)\right)$

Substitute these points into the differentiated equation:

$\begin{aligned}(1,0)\implies\frac{\text{d}y}{\text{d}x}&=(12(1)(0)^3+4(1)(0)^2)/(1-18(1)^2(0)^2-4(1)^2(0))\\\\&=(0)/(0)\\\\&=\sf unde\:\!fined\end{aligned}$

$\begin{aligned}\left(1,(-1-√(7))/(6)\right)\implies\frac{\text{d}y}{\text{d}x}&=(12(1)\left((-1-√(7))/(6)\right)^3+4(1)\left((-1-√(7))/(6)\right)^2)/(1-18(1)^2\left((-1-√(7))/(6)\right)^2-4(1)^2\left((-1-√(7))/(6)\right))\\\\\frac{\text{d}y}{\text{d}x}&=(√(7))/(7)\end{aligned}$

$\begin{aligned}\left(1,(-1+√(7))/(6)\right) \implies \frac{\text{d}y}{\text{d}x}&=(12(1)\left((-1+√(7))/(6)\right)^3+4(1)\left((-1+√(7))/(6)\right)^2)/(1-18(1)^2\left((-1+√(7))/(6)\right)^2-4(1)^2\left((-1+√(7))/(6)\right))\\\\\frac{\text{d}y}{\text{d}x}&=-(√(7))/(7)\end{aligned}$