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Trig Differentiation With Respect To y

Given that

x = cos^2(y^2)

find

dy/dx

(idk how to do fractions with latex) in terms of

y

___________________

What I've done is, differentiate with respect to y to get

dx/dy = -2y*sin(y^2)

And then can I just take the reciprocal so...?

dy/dx = 1/(-2y*sin(y^2))

Not sure if I'm able to do this

(edited 6 years ago)

Reply 1

have

Original post by Retsek

Given that

x = cos^2(y^2)

find

dy/dx

(idk how to do fractions with latex) in terms of

y

___________________

What I've done is, differentiate with respect to y to get

dx/dy = -2y*sin(y^2)

And then can I just take the reciprocal so...?

dy/dx = 1/(-2y*sin(y^2))

Not sure if I'm able to do this

Your method is almost right. It's just that you did it for x= cos(y^2) and not x = cos^2(y^2). I would suggest doing chain rule a second time.

Also to make fractions in latex you do \dfrac{numerator}{denominator}

Reply 2

Kallisto

Entertainment Forum Helper

Original post by Retsek

Given that

x = cos^2(y^2)

find

dy/dx

(idk how to do fractions with latex) in terms of

y

___________________

What I've done is, differentiate with respect to y to get

dx/dy = -2y*sin(y^2)

And then can I just take the reciprocal so...?

dy/dx = 1/(-2y*sin(y^2))

Not sure if I'm able to do this

You did a mistake in your differentiation: if [tex] x = cos²(y²) is your term, it is dy/dx=-4cos(y²) * sin(y²). If you do the reciprocal now, you get dy/dx = 1/-4cos(y²)*sin(y²).

You have forgotten the square in cosine!