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How do you implicitly differentiate #4= xytan(x^2y) #?

Answer 1

Aurora Alvarado

#y'=(-y* tan(x^2y)-2x^2y^2*sec^2(x^2 y))/(x* tan(x^2 y)+x^3 y* sec^2(x^2 y))#

from the given

#4=xy tan(x^2 y)#

differentiate both sides of the equation

#d/dx(4)=d/dx(x)*y*tan(x^2 y)+d/dx(y)*x*tan(x^2 y)+#

#xy*d/dx(tan(x^2 y)#

#0=y*tan(x^2 y)+x*tan(x^2 y) y' + 2x^2 y^2 *sec^2(x^2 y) + x^3 y##sec^2(x^2 y) y'#

#-y tan(x^2 y)-2 x^2 y^2 sec^2 (x^2 y) = (x tan (x^2 y) + x^3 y sec^2(x^2 y)) y'#

#y'=(-y* tan(x^2y)-2x^2y^2*sec^2(x^2 y))/(x* tan(x^2 y)+x^3 y* sec^2(x^2 y))#

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Answer 2

Elijah Abram

To implicitly differentiate the equation (4 = xy\tan(x^2y)), you would apply the chain rule and product rule.

Differentiating both sides with respect to (x):

[0 = y\tan(x^2y) + xy\sec^2(x^2y)(2xy + x^2)]

Then, you can solve for (\frac{dy}{dx}).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.