Question

Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

y sin 12x = x cos 2y, (π/2, π/4)

Answer 1

take derivative both sides
\[y'sin12x +12 y cos 12x=cos2y -2x sin2y\]
\[y'=\frac{cos2y-2xsin2y-12ycos12x}{sin12x}\]
\[at~(\frac{\pi}{2},\frac{\pi}{4})\]
\[y'= \frac{cos 2\frac{\pi}{4}-2\frac{\pi}{2}sin2\frac{\pi}{4}-12\frac{\pi}{4}cos12\frac{\pi}{2}}{sin12\frac{\pi}{2}}\]
\[y'= \frac{cos\frac{\pi}{2}-\pi sin\frac{\pi}{2}-3\pi cos6\pi}{sin6\pi}\]
however, sin6pi =0 therefore y' at that point is undefined or it's vertical line x = pi/2
that's what I think. check please

Answer 2

\( y'sin12x +12 y cos 12x=cos2y - \color{red}{2x sin2y} \)
When you took the derivative here, it seems you did not take the chain rule with the y.
d/dx (x cos (2y)) = cos 2y + 2x y' sin 2y <-- The y' due to taking the derivative of the interior. d/dx(y) =/= 1

Answer 3

@AccessDenied hihihi.. don't get!!

Answer 4

LHS of original one is ysin12x, I take implicit derivative and get y'sin12x +12ycos 12x. that included chain rule, is it not right?

Answer 5

This is concerning the RHS:
\( x \cos 2y \)
Taking the derivative of this, the product rule was correct, but the second term was:
\( \color{#aa0000}{2}x \sin 2y \)
But \(2\) is not the entire derivative of the interior. \( \frac{d}{dx} (2y) = 2y' \)

Answer 6

yes, you are right!! my bad,

Answer 7

ok, let my take off my wrong solution. @TachiHere my bad, I am sorry.

Answer 8

So, we want to find the equation of the tangent line to the curve:
\( y \sin 12x = x \cos 2y \) at the point \( \displaystyle ( \frac{\pi}{2} , \; \frac{\pi}{4} ) \)

We use implicit differentiation by taking the derivative of both sides of the equation, preserving equality:
\( \displaystyle \frac{d}{dx} \left( y \sin 12 x \right) = \frac{d}{dx} \left( x \cos 2y \right) \)
On both sides we may apply the product rule.
\( \displaystyle \frac{d}{dx} (y) \sin 12x + y \frac{d}{dx} \left( \sin 12 x \right) = \frac{d}{dx} (x) \cos 2y + x \frac{d}{dx} \left( \cos 2y \right) \)
We may apply the chain rule to the derivatives of the trigonometric functions, and likewise use the power rule on the powers of x.
\( \displaystyle \frac{dy}{dx} \sin 12x + y \; 12 \cos 12 x = \cos 2y + x \; 2 \frac{dy}{dx} \frac{d (\cos 2y)}{dy} \)
\( \displaystyle y' \sin 12x + 12y \cos 12 x = \cos 2y - 2x y' \sin 2y \)
We may now solve for y' through algebraic manipulations. This will allow us to substitute our point to find the slope of the line at this point.
\( \displaystyle y' \sin 12x + 2x y' \sin 2y = \cos 2y - 12 y \cos 12 x \)
\( \displaystyle y' \left( \sin 12x + 2x \sin 2y \right) = \cos 2y - 12y \cos 12 x \)
\( \displaystyle y' = \frac{\cos 2y - 12 y \cos 12 x}{\sin 12x + 2x \sin 2y} \) Substitute \( \displaystyle ( \frac{\pi}{2} , \; \frac{\pi}{4} ) \)
\( \displaystyle y' = \frac{\cos (2 \frac{\pi}{4}) - 12 \frac{\pi}{4} \cos (12 \frac{\pi}{2})}{\sin (12 \frac{\pi}{2}) + 2 \frac{\pi}{2} \sin (2 \frac{\pi}{4})} \)
\( \displaystyle y' = \frac{\cos \frac{\pi}{2} - 3 \pi \cos 6\pi}{\sin 6 \pi + \pi \sin \frac{\pi}{2}} \)
Simplify the trigonometric expressions:
\( \displaystyle y' = \frac{0 - 3 \pi}{0 + \pi} \)
\( \displaystyle y' = \frac{-3 \pi}{\pi} \)
\( \displaystyle y' = -3 \)

The slope of the line at this point is \(m = -3\). We may then use point-slope to give the equation of the line at this point.
\( \displaystyle y - y_1 = m \left(x - x_1\right) \) ; \( \displaystyle (x_1, \; y_1) = ( \frac{\pi}{2} , \; \frac{\pi}{4} ) \) , \(m = -3 \)
\( \displaystyle y - \frac{\pi}{4} = -3 \left( x - \frac{\pi}{2} \right) \)
This is the equation in point-slope form, although we may modify it to obtain slope-intercept form as well by going a tad further with multiplying the -3 through and simplifying.

A quick check through wolfram:
http://www.wolframalpha.com/input/?i=plot+y+sin+12x+%3D+x+cos+2y%2C++y+%3D+-3%28x+-+pi%2F2%29+%2B+pi%2F4+from+x%3D+pi%2F4+to+pi%2C+y+%3D+pi%2F8+to+pi%2F2
Looks good.

Answer 9

thanks @AccessDenied

Answer 10

Thank you both for the help. I was really lost!

Answer 11

You're welcome! Glad to help! :)