Implicit Differentiation: find dy/dx for x = tan(y)

[catalle]

so I am trying to find \(\displaystyle \frac {dy}{dx}\) for \(\displaystyle x=tan(y)\)
here's some of my work
not sure if I'm right but I'm applying the product rule to find it

\(\displaystyle u=tan\)

\(\displaystyle u'=sec^2\)

\(\displaystyle v=y\)

\(\displaystyle v'= \frac {dy}{dx}\)

\(\displaystyle tan \frac {dy}{dx} + sec^2 y=1\)

the 1 is from the derivative of x

then I subtracted \(\displaystyle sec^2 y\) from both sides

which is where I got stuck
the answer for the problem is supposed to be \(\displaystyle cos^2 y\)
did I do these steps right or no?

 

[galactus]

Re: Implicit Differentiation help

\(\displaystyle x-tan(y)=0\)

\(\displaystyle 1-sec^{2}(y)y'=0\)

\(\displaystyle y'=\frac{1}{sec^{2}(y)}=cos^{2}(y)\)

That's about it.

 

[Deleted member 4993]

Re: Implicit Differentiation help

\(\displaystyle x-tan(y)=0\)

\(\displaystyle 1-sec^{2}(y)y'=0\)

\(\displaystyle y'=\frac{1}{sec^{2}(y)}=cos^{2}(y)\)

You can go little further

\(\displaystyle tan (y) \, = \, \frac{x}{1}\)

\(\displaystyle cos(y) \, = \, \frac{1}{\sqrt{1+x^2}}\)

\(\displaystyle y' \, = \, \frac{1}{1+x^2}\)

Thus

\(\displaystyle \frac{d}{dx}[tan^{-1}(x)] \, =\, \frac{1}{1+x^2}\)

That's about it.

 

[catalle]

Re: Implicit Differentiation help

\(\displaystyle x-tan(y)=0\)

\(\displaystyle 1-sec^{2}(y)y'=0\)

\(\displaystyle y'=\frac{1}{sec^{2}(y)}=cos^{2}(y)\)

That's about it.

could you please explain what you did?
how those steps could fit in with my steps?

 

[catalle]

Re: Implicit Differentiation help

\(\displaystyle x-tan(y)=0\)

\(\displaystyle 1-sec^{2}(y)y'=0\)

\(\displaystyle y'=\frac{1}{sec^{2}(y)}=cos^{2}(y)\)

You can go little further

\(\displaystyle tan (y) \, = \, \frac{x}{1}\)

\(\displaystyle cos(y) \, = \, \frac{1}{\sqrt{1+x^2}}\)

\(\displaystyle y' \, = \, \frac{1}{1+x^2}\)

Thus

\(\displaystyle \frac{d}{dx}[tan^{-1}(x)] \, =\, \frac{1}{1+x^2}\)

That's about it.

that is really specific
we haven't learned about inverse trig functions yet
though it is the next section in the book...

 

[Deleted member 4993]

Re: Implicit Differentiation help

\(\displaystyle x-tan(y)=0\)

\(\displaystyle 1-sec^{2}(y)y'=0\)

\(\displaystyle y'=\frac{1}{sec^{2}(y)}=cos^{2}(y)\)

That's about it.

could you please explain what you did? <<< Simply ISOLATE y' - and solve for it.

how those steps could fit in with my steps?