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$$x^2+xy+y^2=3, (1,1)$$ I got the derivative as.. $$\frac{2x-2}{x+4}$$ But when I plug in the points I get the equation $y=x/2+2$ which is wrong. Is my derivative wrong? Or am I making a mistake plugging my numbers in. If you could show me where I'm going wrong it'd be much appreciated.

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3 Answers 3

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I'm getting a very different derivative. Here's how I did it

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

Now we differentiate

$$0=2x+x \frac{dy}{dx}+y+2y \frac{dy}{dx} $$

Separate terms

$$-(2x+y)=\frac{dy}{dx}(x+2y)$$

so

$$\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}$$

Plugging in $(1,1)$, we get $\frac{dy}{dx}=\frac{-(2+1)}{1+2}=\frac{-3}{3}=-1$, so the tangent line at $(1,1)$, in point-slope form, is $y-1=-(x-1)$, or $y=-x+2$

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$$x^2+xy+y^2=3\\ 2x+x\frac{dy}{dx}+y+2y\frac{dy}{dx}=0\\ \frac{dy}{dx}=-\frac{2x+y}{x+2y}$$ At $(1,1)$, $\frac{dy}{dx}=-1$. So, $y-1=-(x-1)=1-x$. So, $y=2-x$ is the equation for the tangent line.

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Ok what about after finding the first derivative I make y the subject in the main function and then substitute the x(1) to find the gradient of the tangent Please i want a clarification

solution

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    $\begingroup$ It's not really clear if this is an answer to the question, or a new question. If you meant to do the latter, please use the Ask Question button. $\endgroup$
    – Glorfindel
    Apr 25, 2017 at 14:03

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