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About Scotto
Expertise
Any kind of mathematics (calculus, analysis, game theory, linear approximation, finite differences, linear regression, linear programming, numerical analysis, probability, statistics, etc.). I also have answered some questions in Physics (mass, momentum, falling bodies), Chemistry (charge, reactions, symbols, molecules), and Biology.

Experience
Experience in the area: I have tutored students in all areas of mathematics for over 20 years. Education/Credentials: BSand MS in Mathematics from Oregon State University, where I completed sophomore course in Physics and Chemistry. I received both degrees with high honors. Awards and Honors: I have passed Actuarial tests 100, 110, and 135.

Publications
Maybe not a publication, but I have respond to well oveer 3000 questions on the PC. That's around 2,000 in basic math and 1,000 in advanced math.

Education/Credentials
I aquired well over 40 hours of upper division courses. This was well over the number that were required. I graduated with honors in both my BS and MS degree from Oregon State University. I was allowed to jump into a few junior level courses my sophomore year.

Awards and Honors
I have been nominated as the expert of the month several times. All of my scores right now are at least a 9.8 average (out of 10).

Past/Present Clients
My past clients have been students at OSU, students at the college in South Seattle, referals from a company, friends and aquantenances, people from my church, and people like you.

You are here: Experts > Teens > Homework/Study Tips > Calculus > Slope of line tangent to curve

Calculus - Slope of line tangent to curve

Expert: Scotto - 11/4/2009

Question
The slope of the line tangent to the curve y squared + (xy +1)cubed= 0 at (2,-1) is...

Answer
Let x be a function on y, so x is f(y).
That is, y² + (xy + 1)³ = 0.
This says that (xy+1)³ = -y².
Taking the cuberoot of both sides gives xy + 1 = cr((-y²)^(1/3)), were cr is for cube-root.
Subtract 1 from both sides and then divide both sides by y.
This gives x = (cr((-y²)^(1/3)) - 1)/y = g(y)/y where g(y) = cr((-y²)^(1/3)) - 1.

The chain rule says that dx/dy = (lo d hi - hi d lo)/lo² = (yg'(y) - g(y)(1))y².

We were given g(y) = cr((-y²)^(1/3)) - 1,
so g'(y) = (1/3)/cr(-y^(1/3)) = 1/(3cr(-y^(1/3)).

Put in -1 for y and get -1/slope of the curve.

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