MHB Josh's question at Yahoo Answers regarding implicit differentiation

AI Thread Summary
The discussion centers on deriving the implicit differentiation of the equation 2x + y² - sin(xy) = 0. The method involves applying the chain rule and product rule to differentiate with respect to x. The resulting expression for dy/dx is derived as (y cos(xy) - 2) / (2y - x cos(xy)). The response encourages further calculus questions in the forum. This detailed approach provides a clear solution to the implicit differentiation problem presented by Josh.
MarkFL
Gold Member
MHB
Messages
13,284
Reaction score
12
Here is the question:

Derive Trig Function?

2. 2x + y2 – sin(xy) = 0

Please help me derive this, thanks!Additional Details

the equation is 2x + y^2 - sin(xy) = 0 The y is SQUARED

Here is a link to the question:

Derive Trig Function? - Yahoo! Answers

I have posted a link there to this topic so the OP can find my response.
 
Mathematics news on Phys.org
Hello Josh,

We are given to find $$\frac{dy}{dx}$$ for the implicit relation:

$$2x+y^2-\sin(xy)=0$$

So, we may use implicit differentiation since the equation cannot be solved for either $x$ or $y$. This involves using the chain rule, along with the other rules for differentiation.

Hence, we find by implicitly differentiating with respect to $x$:

$$2+2y\frac{dy}{dx}-\cos(xy)\frac{d}{dx}(xy)=0$$

Applying the product rule, there results:

$$2+2y\frac{dy}{dx}-\cos(xy)\left(x\frac{dy}{dx}+y \right)=0$$

Distributing the cosine function, we have:

$$2+2y\frac{dy}{dx}-x\frac{dy}{dx}\cos(xy)-y\cos(xy)=0$$

Now, we move all terms not having $$\frac{dy}{dx}$$ as a factor to the right side, and factor out $$\frac{dy}{dx}$$ on the left side:

$$\frac{dy}{dx}\left(2y-x\cos(xy) \right)=y\cos(xy)-2$$

Divide through by $$2y-x\cos(xy)$$ to obtain:

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

To Josh and any other guests viewing this topic, I invite and encourage you to post other calculus problems here in our Calculus forum.

Best Regards,

Mark.
 
Seemingly by some mathematical coincidence, a hexagon of sides 2,2,7,7, 11, and 11 can be inscribed in a circle of radius 7. The other day I saw a math problem on line, which they said came from a Polish Olympiad, where you compute the length x of the 3rd side which is the same as the radius, so that the sides of length 2,x, and 11 are inscribed on the arc of a semi-circle. The law of cosines applied twice gives the answer for x of exactly 7, but the arithmetic is so complex that the...
Is it possible to arrange six pencils such that each one touches the other five? If so, how? This is an adaption of a Martin Gardner puzzle only I changed it from cigarettes to pencils and left out the clues because PF folks don’t need clues. From the book “My Best Mathematical and Logic Puzzles”. Dover, 1994.
Back
Top