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Greetings and salutations.
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In this question, we're given this curve, y squared minus 5x squared plus y minus 4x minus 3 is equal to 0.
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We're asked various things.
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The first is to use implicit differentiation to find dy dx.
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So when we're using implicit differentiation, essentially we want to go ahead and take the derivative of this entire equation with respect to x.
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What does that look like? well for this first term here dy squared dx, essentially we take the derivative of whatever the term is as normal.
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So the derivative of y squared is going to be 2y, but then we're differentiating with respect to x.
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So then we have to technically do a chain rule here, and the chain rule, well our variable here is dy, that's our variable, and we're differentiating that with respect to x.
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So we'll end up with 2y dy dx.
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For this second term here, we're going to have that minus 5, and then we take the derivative of this term, the derivative of x squared, of course, is going to be 2x.
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And then technically what we're doing is we're saying we're taking the derivative with respect to x.
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What was our variable? it was x.
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Well, dx dx is just 1, and that's why we generally don't write it.
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Third term, we want to take the derivative of y.
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Well, the derivative of y is going to be 1 times d dx of whatever our variable was, which was y, minus 4 times the derivative of of x is going to be 1, and then we're going to have dx dx there, and then minus 0, because the derivative of constant with respect to anything is always going to be 0.
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So these dx dx's obviously go away, and we'll be left with 2y dy dx minus 10x plus dy dx minus 4 is equal to 0.
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We then non -dy dx terms to the other side and then factoring out dy dx and dividing through.
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And so we'll end up with this term here for dy dx.
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For part b we're asked verify that the point 1 3 is on the graph of the curve.
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So this equation here is the graph of the curve.
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Go ahead and get rid of these parentheses.
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And what we want to do is we just want to verify that the point 1 3 does in fact satisfy this equation.
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So we want to check the point 1, 3 by just plugging it into here...