I dont get it when it means "to find $dy/dx$ in terms of $x$ and $y$" what does it mean? Solve the following problem and write your answer and solutions on bond paper.
1) Use implicit differentiation to find $dy/dx$ in terms of $x$ and $y$.
a) $2x^3=2y^2+5$
b) $1=3x+2x^2y^2$
c)$x^2=(4x^2y^3+1)^2$
2) Use implicit differentiation to find second derivative in terms of $x$ and $y$.
$4y^2+2=3x^2$
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$\begingroup$We need to express the derivative of the functions in terms of both $x$ and $y$ (in our answers), rather than just $x$. This is how it works:
$$a) 2x^3 = 2y^2+5$$$$\frac{d}{dx} 2x^3 = \frac{d}{dx}2y^2 + \frac{d}{dx}5$$$$6x = \frac{d}{dx}2y^2+0 \implies 6x = \frac{d}{dx}2y^2$$Here, we need to use the Chain Rule to simplify $\frac{d}{dx}2y^2$. $$\frac{du}{dx} = \frac{du}{dy}\cdot\frac{dy}{dx}$$$$\frac{d}{dx}2y^2 = \frac{d}{dy}2y^2\cdot\frac{dy}{dx}$$$$\frac{d}{dx}2y^2 = 4y\cdot\frac{dy}{dx}$$Now, we can continue.
$$6x = 4y\cdot\frac{dy}{dx}$$$$\frac{dy}{dx} = \frac{6x}{4y} \implies \frac{dy}{dx} = \frac{3x}{2y}$$We used the derivatives of both sides in order to differentiate implicitly. As you can see, it is expressed in terms of both x and y, just as the question asked. Repeating this process for the other questions will lead to the answer.$$b) 1= 3x+2x^2y^2$$$$\frac{d}{dx}1 = \frac{d}{dx}3x+\frac{d}{dx}2x^2y^2$$$$0 = 3+\frac{d}{dx}2x^2y^2$$$$-3 = \frac{d}{dx}2x^2y^2$$To find $\frac{d}{dx}2x^2y^2$, use the Product Rule.$$\frac{d}{dx}2x^2y^2 = (\frac{d}{dx}2x^2)\cdot y^2+(\frac{d}{dx}y^2)\cdot 2x^2 \implies \frac{d}{dx}2x^2y^2 = 4xy^2+2x^2\cdot 2y\cdot \frac{dy}{dx}$$$$\frac{d}{dx}2x^2y^2 = 4xy^2+4x^2y\cdot\frac{dy}{dx}$$ Back to the question, we can quickly reach the answer.$$-3 = 4x^2y+4xy^2\cdot\frac{dy}{dx} \implies -3-4x^2y = 4xy^2\cdot\frac{dy}{dx} \implies \frac{dy}{dx} = \frac{-3-4x^2y}{4xy^2} = -\frac{3+4x^2y}{4xy^2}$$$$c) x^2 = (4x^{2}y^{3}+1)^2$$$$\frac{d}{dx}x^2 = \frac{d}{dx}(4x^{2}y^{3}+1)^2$$$$2x = \frac{d}{dx}(4x^{2}y^{3}+1)^2$$Now, $\frac{d}{dx}(4x^{2}y^{3}+1)^2$ should be calculated.
Using the Chain Rule, $$\frac{dy}{dx} = \frac{dy}{du}\cdot\frac{du}{dx}$$$$y = x^2 = (4x^{2}y^{3}+1)^2$$ $$u = (4x^{2}y^{3}+1)$$$$\frac{d}{dx}(4x^{2}y^{3}+1)^2 = 2(4x^{2}y^{3}+1)\cdot\frac{d}{dx}(4x^{2}y^{3}+1)$$$$\frac{d}{dx}(4x^{2}y^{3}+1)^2 = 2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})\cdot\frac{dy}{dx}$$Back to the question.$$2x = 2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})\cdot\frac{dy}{dx}$$$$\implies \frac{dy}{dx} = \frac{2x}{2(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})}$$ $$\implies \frac{dy}{dx} = \frac{x}{(4x^{2}y^{3}+1)\cdot (8xy^3+12x^{2}y^{2})}$$For question 2, the process is repeated twice with substitution at the end.$$2) 4y^2+2 = 3x^2$$$$\frac{d}{dx}4y^2+\frac{d}{dx}2 = \frac{d}{dx}3x^2$$$$\frac{d}{dx}4y^2 + 0 = 6x \implies \frac{d}{dx}4y^2 = 6x$$Using the Chain Rule, we get $$8y\cdot\frac{dy}{dx} = 6x \implies \frac{dy}{dx} = \frac{6x}{8y} = \frac{3x}{4y}$$But the question wanted the double derivative, so the process repeats.$$\frac{dy}{dx}\frac{3x}{4y} = \frac{\frac{d}{dx}(3x)\cdot 4y -\frac{d}{dx}(4y)\cdot 3x}{(4y)^2}$$ $$\frac{d}{dx}\frac{3x}{4y} = \frac{12y-12x\cdot\frac{dy}{dx}}{16y^2}$$Now, plug in $\frac{3x}{4y}$ for $\frac{dy}{dx}$. $$\frac{d}{dx}\frac{3x}{4y} = \frac{12y-12x\cdot\frac{3x}{4y}}{16y^2}$$Simplify the right side and you’ll get this.$$\frac{d}{dx}\frac{3x}{4y} = \frac{12y^2-9x^2}{16y^3}$$Therefore, this is the second derivative. $$\frac{d^2y}{dx^2} = \frac{12y^2-9x^2}{16y^3}$$
It means that in your answer you are required to have only x and y as independent variables*, and not $y'$ or $\frac{dy}{dx}$ or anything else.
As far as I understand, with the first derivative it can't be done other way, since you don't introduce any other variables into the equation, like here, from your example:
$$ 2x^2=2y^2+5 \to 4x = 4yy' \to y' = \frac{x}{y} $$
Here the answer contains only x and y, which means exactly "in terms of x and y".
When one needs the second derivative, this is where you can have another variable, namely the first derivative. So you are required not to use it in your final answer, which usually means to substitute it with the first derivative equation (which contains only x and y)
From your question: Use implicit differentiation to find second derivative in terms of x and y:
$$ 4y^2+2=3x^2 \\ 8yy' = 6x \\ 4yy' = 3x \\ y' = \frac{3x}{4y} $$
this is the first derivative.
The second one is:
$$ 4yy' = 3x \\ 4y'y' + 4yy''=3 \\ 4yy'' = 3 - 4(y')^2\\ y'' = \frac{3 - 4(y')^2}{4y} $$
This is the second derivative, but it's not "in terms of x and y" since it contains $y'$ as an independent variable. You just need to substitute it with the value from $y' = \frac{3x}{4y}$, that is:
$$ y'' = \frac{3 - 4(\frac{3x}{4y})^2}{4y}\\ y'' = \frac{3}{4y} - \frac{9x^2}{16y^3}\\ $$
Here the equation contains only x and y as independent variables, which means it is in terms of x and y.
* and independent variable is one, which can be freely substituted for any allowed value, unlike the dependent one, which must be calculated from values of the independent variables.
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