Find dy/dx by implicit differentiation for #sqrt(5x + y)= 3 + x^2y^2# ?

Answer 1

#y'=(4xy^2sqrt(5x+y)-5)/(1-x^2ysqrt(5x+y))#

Using the chain rule by assuming #y=y(x)# we get
#1/2(5x+y)^(-1/2)(5+y')=2xy^2+2x^2yy'#
multiplying both sides by #2sqrt(5x+y)# we obtain
#5+y'=4xy^2sqrt(5x+y)+4x^2yy'sqrt(5x+y)# solving for #y'#
#y'(1-4x^2ysqrt(5x+y))=4x^2ysqrt(5x+y)-5#

so

#y'=(4xy^2sqrt(5x+y)-5)/(1-4x^2ysqrt(5x+y))#
if #1-4x^2ysqrt(5x+y)ne 0#
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Answer 2

Sure, let's find ( \frac{dy}{dx} ) using implicit differentiation for the equation ( \sqrt{5x + y} = 3 + x^2y^2 ).

First, we differentiate both sides of the equation with respect to ( x ) term by term:

For the left side, we apply the chain rule: [ \frac{d}{dx}(\sqrt{5x + y}) = \frac{1}{2\sqrt{5x + y}} \cdot \frac{d}{dx}(5x + y) = \frac{1}{2\sqrt{5x + y}} \cdot (5 + \frac{dy}{dx}) ]

For the right side: [ \frac{d}{dx}(3 + x^2y^2) = 0 + 2xy^2 \frac{dx}{dx} + x^2(2yy') ]

Now, equating the derivatives: [ \frac{1}{2\sqrt{5x + y}} \cdot (5 + \frac{dy}{dx}) = 2xy^2 + 2x^2yy' ]

Solve for ( \frac{dy}{dx} ): [ \frac{dy}{dx} = 4xy^2 + 4x^2yy' - \frac{5}{2\sqrt{5x + y}} ]

This is the implicit derivative of ( y ) with respect to ( x ).

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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.

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.

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.

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.

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