 In this video, we provide the solution to question number 13 from Practice Exam 3 for Math 1210. We are asked to find the equation of the line tangents to the curve 3x squared plus 17 sine of y equals 3 plus xy at the point 1 comma 0. We need to write our line, the tangent line in slope intercept form. So the first thing we need to do is we need to find the derivative. We need to find dy over dx at the point 1 comma 0. So we're going to find the derivative implicitly. So we're going to take the derivative of both sides, 3x squared plus 17 sine of y on the left hand side, the right hand side. It's going to be 3 plus xy take the derivative. Now our goal is to find the derivative dy over dx. So we are taking the derivative with respect to x. Keep that in mind as we go through this calculation. So when you take the derivative, we have to take derivative 3x squared. We have to take the derivative of the 17 sine of y. And you can do all of this in the m1 line. We don't necessarily have to do it step by step by step. We can't just jump to the end. We need to show our work. But we need to show enough work that a typical student in calculus can read what you wrote down and understand. If your jumps are too large for the typical student to understand, that means you've taken two big steps here. So the typical student could understand that the derivative of 3x squared will be a 6x. The typical student could also understand that the derivative of 17 sine of y would be 17 cosine of yy prime. The most common mistake we make when we do implicit differentiations, we forget the inner derivative. We forget the y prime. We need that y prime because we're trying to compute y prime at this moment. The derivative of 3 on the right hand side is 0. And then we have to take the derivative of xy, which the product rule comes into play. We could x prime y plus xy prime like so, but x prime itself is just 1. So if we had scratched it out, that's okay because the derivative there is just a 1. All right, now at this moment we need to solve for y prime. So we could solve for y prime purely algebraically, but we also know that for the slope, we actually care about x equals 1 and y equals 0. So we could actually make those substitutions in here. x equals 1, x equals 1, we see that. And we see that y should equal 0. So we get that here and we get that here. So with those simplifications, we're going to get 6 times 1 plus 17 times cosine of 0 times y prime. This is equal to 0 plus 1 times y prime. That simplifies things dramatically. 6 times 1 is 6. Cosine of 0 is actually 1. So you get 6 plus 17 y prime. And on the right hand side, you just get a y prime. So we need to solve for y prime here. Subtract 17 y prime from both sides. That would then give us that 6 is equal to negative 16 y prime. Divide both sides by negative 16. We get that dy over dx is going to equal 6 over negative 16. But there's a common factor of 2 so you can simplify it to be negative 3 over 8. Don't panic too much if you get a fraction because slopes are generally fractions, right? Rise over 1. It's perfectly fine. Now what we want to do is we actually want to come up with our tangent line. The tangent line is going to look like y minus f of a here is equal to f prime at a times x minus a. Where a and f of a are the point of tangency. So we get y minus, the y coordinate here is 0. The derivative, the slope of the tangent line is negative 3 eighths. And then we get x minus 1 right here. Distribute the negative 3 eighths. We get y is equal to negative 3 eighths x plus 3 eighths. And so this is then our correct tangent line equation written in slope intercept form.