jiujuan
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CALC BC HELPPP!!!??? 100PTS!!!

4x2+4x+xy=5 and y(5)=−23, find y′(5) by implicit differentiation.

Sagot :

the assumption being that "x" is a plain variable whilst "y" is a function, that matters because the chain rule would be needed for a function, not so for a plain variable.

[tex]4x^2+4x+xy=5\implies 8x+4+\stackrel{\textit{product rule}}{\left( 1\cdot y+x\cdot \cfrac{dy}{dx} \right)}=0 \\\\\\ x\cfrac{dy}{dx}=-8x-4-y\implies \cfrac{dy}{dx}=\cfrac{-8x-4-y}{x}[/tex]

now, we know that y(5) = -23, which is another way of saying that when x = 5, y = -23, but we already knew that, we can get that by simply plugging it into the equation hmmm y'(5), well

[tex]\left. \cfrac{dy}{dx}=\cfrac{-8x-4-y}{x} \right|_{\stackrel{x=5~}{\textit{\tiny y=-23}}}\implies \cfrac{-8(5)-4-(-23)}{5}\implies \cfrac{-21}{5}[/tex]

semsee45

Answer:

[tex]\dfrac{dy}{dx}=-\dfrac{21}{5}[/tex]

Step-by-step explanation:

Given function:

[tex]4x^2+4x+xy=5[/tex]

To differentiate the given function using implicit differentiation:

[tex]\dfrac{d}{dx}4x^2+\dfrac{d}{dx}4x+\dfrac{d}{dx}xy=\dfrac{d}{dx}5[/tex]

Differentiate terms in x only (and constant terms) with respect to x:

[tex]\implies 8x+4+\dfrac{d}{dx}xy=0[/tex]

Use the product rule to differentiate the term in x and y:

[tex]\textsf{let }\: u = x \implies \dfrac{du}{dx} = 1[/tex]

[tex]\textsf{let }\: v = y \implies \dfrac{dv}{dx} = \dfrac{dy}{dx}[/tex]

[tex]\begin{aligned}\implies \dfrac{dy}{dx} & =u \dfrac{dv}{dx}+v\dfrac{du}{dx}\\ & =x\dfrac{dy}{dx}+y\end{aligned}[/tex]

[tex]\implies 8x+4+x\dfrac{dy}{dx}+y=0[/tex]

Rearrange to make dy/dx the subject:

[tex]\implies x\dfrac{dy}{dx}=-8x-y-4[/tex]

[tex]\implies \dfrac{dy}{dx}=-\dfrac{8x+y+4}{x}[/tex]

When x = 5, y = -23.  Therefore, substitute these values into the differentiated function to find y'(5):

[tex]\implies \dfrac{dy}{dx}=-\dfrac{8(5)+(-23)+4}{5}[/tex]

[tex]\implies \dfrac{dy}{dx}=-\dfrac{21}{5}[/tex]

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