Question

Find dy/dx: x= radical of tan (y^2).

Answer 1

Need help with this question: \[x= \sqrt{\tan (y ^{2)}}\] anyone help please!!

Answer 2

Isolate y first.

So y = \[y = \sqrt{\tan^{-1} (x^2)}\]

Then take the derivative with respect to x.

Answer 3

Can anyone go through this whole method with me till the end bcoz i am having tought time understanding this problem.

Answer 4

Sure :)

Answer 5

\[\large x=\sqrt{\tan y^2}\]
If you rewrite the square root as a rational expression it might make it easier to work with. If you have the derivative of sqrt memorized by this point, then you can skip this step.\[\large x=(\tan y^2)^{1/2}\]

Taking the derivative of both sides WRT (with respect to) x gives us,\[\large 1=\color{orangered}{\frac{1}{2}\left(\tan y^2\right)^{-1/2}}\color{royalblue}{\left(\tan y^2\right)'}\]

So what we did there was,
On the left, we took the derivative of x, giving us 1.
On the right, we took the derivative of the OUTERMOST function, the 1/2 power.
Applying the power rule gave us the Orange part.

Then we have to also apply the chain rule, multiplying by the derivative of the inside. The prime is to show that we still need to take the derivative of that part.

Answer 6

oh wow thanks you described very well :)

Answer 7

We still have some work to do! Simmer down! :D

Answer 8

So now we have to take care of the blue part.

Answer 9

ahan ok

Answer 10

Taking the derivative of the blue part gives us this new blue part.
\[\large 1=\frac{1}{2}\left(\tan y^2\right)^{-1/2}\color{royalblue}{\left(\sec^2 y^2\right)}\color{#33B111}{(y^2)'}\]

But once again we have to apply the chain rule, multiplying by the derivative of the inside.

Answer 11

Now we'll take the derivative of the green part,\[\large 1=\frac{1}{2}\left(\tan y^2\right)^{-1/2}\left(\sec^2 y^2\right)\color{#33B111}{(2y)}\color{#662FFF}{(y)'}\]Which gives us this new green part, and then we have to apply the chain rule AGAINNNNNNNNNN.

Answer 12

We've made it all the way down to the \(y\)!
Taking the derivative of \(y\) gives us \(y'\) or \(\dfrac{dy}{dx}\).\[\large 1=\frac{1}{2}\left(\tan y^2\right)^{-1/2}\left(\sec^2 y^2\right)(2y)\dfrac{dy}{dx}\]

Answer 13

Oh boy this is a mean question :3

Answer 14

Now to solve for dy/dx we'll divide both sides by this big purple blob.\[\large 1=\color{#662FFF}{\frac{1}{2}\left(\tan y^2\right)^{-1/2}\left(\sec^2 y^2\right)(2y)}\dfrac{dy}{dx}\]

Answer 15

Oh I guess we should simplify things down first actually...

Answer 16

\[\large 1=\cancel{\frac{1}{2}}\left(\tan y^2\right)^{-1/2}\left(\sec^2 y^2\right)(\cancel{2}y)\dfrac{dy}{dx}\]Which simplifies to,\[\large 1=\color{#662FFF}{\frac{y \sec^2 y^2}{\sqrt{\tan y^2}}}\frac{dy}{dx}\]We'll multiply both sides by the reciprocal of the purple thing,\[\large \frac{dy}{dx}=\frac{\sqrt{\tan y^2}}{y \sec^2 y^2}\]

Answer 17

Wow this is an evil evil problem :D Lemme look over my steps a minute, to make sure I didn't make any silly mistakes.

Answer 18

Nope that turned out correctly it seems.
Oh I guess we can convert the secant to its equivalent in cosine.\[\large \frac{dy}{dx}=\frac{\sqrt{\tan y^2}}{y \sec^2 y^2} \qquad \rightarrow \qquad \large \frac{dy}{dx}=\frac{\sqrt{\tan y^2}}{y \dfrac{1}{\cos^2y^2}}\]Dividing by a fraction, we can rewrite this as multiplication by flipping the bottom fraction,\[\large \frac{dy}{dx}=\frac{\cos^2y^2\sqrt{\tan y^2}}{y}\]

Answer 19

This problem might be easier if you do it the way Canadian suggested...
Isolating \(y\) first. Maybe... I'm not sure though..