Atia H.

# What is tan^3 (xy^2+y) = x?

Calculus: Implicit Differentiation

## 2 Answers By Expert Tutors

By:

Bobosharif S. answered • 02/07/18

Tutor
4.4 (32)

Mathematics/Statistics Tutor

Kenneth S. answered • 02/07/18

Tutor
4.8 (62)

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Atia H.

Oh yes, I forgot to add it. Find dy/dx
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02/07/18

Kenneth S.

Using implicit differentiation (and the chain rule, appropriately):

3 tan2(xy2+y)sec2(xy2+y)(y2y'+y2+y') = 1

Now you can solve for y'
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02/07/18

Arturo O.

Kenneth,

I think that last factor in the left hand side (from the chain rule) should be

(xy2 + y)' = y2 + x(2yy') + y' = y2 + (2xy + 1)y'
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02/07/18

Bobosharif S.

Kenneth, It seems to me that it should be
3 tan2(xy2+y)sec2(xy2+y)(y2 2x y y'+y') = 1.
Below I used differentiation of implicit function's.

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02/07/18

Arturo O.

...with a "+" sign between y2 and 2xyy'.
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02/07/18

Bobosharif S.

Yes, right. Thank you Arturo
3 tan2(xy2+y)sec2(xy2+y)(y2+ 2x y y'+y') = 1.
I did't see your first comment in time :-)

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02/07/18

Kenneth S.

I ACKNOWLEDGE that I screwed up the last parenthesized factor, where the chain rule is applied to the argument of the trig function.  It is (I believe) impossible to correct a Comment here in WyzAnt, in place.
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02/08/18

Atia H.

Thank you all for your replies. Would y^1= 1-3y^2tan^2(xy^2+y)sec^2(xy^2+y) / 3(2xy+1)tan^2(xy^2+y)sec^2(xy^2+y)?
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02/08/18

Bobosharif S.

Yes, and you can simplify it.
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02/08/18

Atia H.

So it would be y^1= 1-3y^2tan^2(xy^2+y)sec^2(xy^2+y) / (6xy+3)tan^2(xy^2+y)sec^2(xy^2+y)?
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02/08/18

Bobosharif S.

You have to find y'=df/dx. From above calculations it follows that
y'=(-3 y2 + cos(y(1 + x y))2 cot(y (1 + x y))2)/(3(1 + 2 x y)). (**)
If you simplify RHS expression in your last comment you get (**)!
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02/08/18

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