### Video Transcript

Find the slope of the tangent to the curve 13π₯ squared plus π¦ squared plus four π₯ minus 20π¦ plus 10 is equal to zero at negative one, one.

The question wants us to find the slope of the tangent line to a curve which is defined implicitly at the point negative one, one. We recall we can find the slope of a tangent line to a curve by finding the derivative of π¦ with respect to π₯. And since we want to find the slope of our tangent at the point negative one, one, we want to find the slope when π₯ is equal to negative one and when π¦ is equal to one.

To find dπ¦ by dπ₯, weβll differentiate both sides of our equation for our curve with respect to π₯. This gives us the derivative of 13π₯ squared plus π¦ squared plus four π₯ minus 20π¦ plus 10 with respect to π₯ is equal to the derivative of zero with respect to π₯. We can differentiate each of these terms separately. First, the derivative of a constant is equal to zero. So the derivative of 10 and the derivative of zero are both equal to zero. Next, we can differentiate both 13π₯ squared and four π₯ by using the power rule for differentiation. We multiply by the exponent and reduce the exponent by one. This gives us 26π₯ and four.

To differentiate our last two terms, we recall that π¦ is a function of π₯. So to differentiate these two terms, weβre going to need to recall some of our rules of implicit differentiation. Since π¦ is a function of π₯, the chain rule tells us the derivative of π of π¦ with respect to π₯ is equal to the derivative of π with respect to π¦ multiplied by the derivative of π¦ with respect to π₯. This gives us the derivative of π¦ squared with respect to π₯ is equal to the derivative of π¦ squared with respect to π¦ multiplied by dπ¦ by dπ₯. And we can evaluate the derivative of π¦ squared with respect to π¦ by using the power rule for differentiation. This gives us two π¦. So weβve shown the derivative of π¦ squared with respect to π₯ is equal to two π¦ times dπ¦ by dπ₯.

So the only term left which we need to differentiate is negative 20π¦. Again, by applying the chain rule, we have the derivative of negative 20π¦ with respect to π₯ is equal to the derivative of negative 20π¦ with respect to π¦ multiplied by dπ¦ by dπ₯. And we can evaluate the derivative of negative 20π¦ with respect to π¦ by using the power rule for differentiation. We get negative 20. So this gives us the derivative of negative 20π¦ with respect to π₯ is equal to negative 20 times dπ¦ by dπ₯. So we now have that 26π₯ plus two π¦ dπ¦ by dπ₯ plus four minus 20 dπ¦ by dπ₯ plus zero is equal to zero. And remember, we want to find the slope of our tangent. So we want to rearrange this equation to make dπ¦ by dπ₯ the subject.

On the left-hand side of our equation, two of our terms share a factor of dπ¦ by dπ₯. So weβll factor this out to get two π¦ minus 20 multiplied by dπ¦ by dπ₯. We want to make dπ¦ by dπ₯ the subject of this equation. So weβll subtract 26π₯ and four from both sides of the equation. So we now have that two π¦ minus 20 times dπ¦ by dπ₯ is equal to negative 26π₯ minus four. Finally, we can make dπ¦ by dπ₯ the subject of this equation by dividing through by two π¦ minus 20. This gives us dπ¦ by dπ₯ is equal to negative 26π₯ minus four all divided by two π¦ minus 20.

Remember, weβre trying to find the slope of our tangent when π₯ is equal to negative one and π¦ is equal to one. So to find the slope of our tangent at negative one, one, we substitute π₯ is equal to negative one and π¦ is equal to one into our equation for dπ¦ by dπ₯. This gives us negative 26 multiplied by negative one minus four all divided by two times one minus 20. Calculating this, we get 22 divided by negative 18. Canceling the shared factor of two in our numerator and our denominator gives us negative 11 divided by nine.

Therefore, weβve shown the slope of the tangent to the curve 13π₯ squared plus π¦ squared plus four π₯ minus 20π¦ plus 10 is equal to zero at the point negative one, one is equal to negative 11 divided by nine.