How do you differentiate #x-cos(x^2)+y^2/x+3x^5=4x^3#?

Answer 1

# dy/dx = (y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2))/(2xy)#

We have:

# x-cos(x^2)+y^2/x+3x^5=4x^3 #

Method 1: Implicit differentiation, as is:

Differentiating wrt #x# whilst applying the chain rule and the product rule:
# 1 + sin(x^2)(d/dx x^2) + ( (x)(d/dxy^2) - (y^2)(d/dx x) ) / (x)^2 + 15x^4 = 12x^2 #
# :. 1 + sin(x^2)(2x) + ( (x)(2ydy/dx) - (y^2)(1) ) / (x)^2 + 15x^4 = 12x^2 #
# :. 1 + 2x sin(x^2) + ( 2xydy/dx - y^2 ) / x^2 + 15x^4 = 12x^2 #
# :. ( 2xydy/dx - y^2 ) / x^2 = 12x^2 - 15x^4 - 1 - 2x sin(x^2)#
# :. 2xydy/dx - y^2 = 12x^4 - 15x^6 -1 - 2x^3 sin(x^2)#
# :. 2xydy/dx = y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2)#
# :. dy/dx = (y^2+12x^4 - 15x^6 -x^2 - 2x^3 sin(x^2))/(2xy)#

Method 2: Implicit Function Theorem

Putting:

# F(x,y) = x-cos(x^2)+y^2/x+3x^5-4x^3 #

We have:

# (partial F)/(partial x) = 1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2 #
# (partial F)/(partial y) = (2y)/x#

Then:

# dy/dx = - ((partial F)/(partial x)) / ((partial F)/(partial y)) # # \ \ \ \ \ \ = - (1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2) / ( (2y)/x ) # # \ \ \ \ \ \ = - ( (x) (1 + 2xsin(x^2) - y^2/x^2 + 15x^4 - 12x^2) ) / (2y) #

Yielding the same result as above.

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Answer 2

To differentiate the given equation (x - \cos(x^2) + \frac{y^2}{x} + 3x^5 = 4x^3) with respect to (x), you can use the rules of differentiation, including the product rule, chain rule, and quotient rule where applicable. After differentiating each term individually, you'll have the derivative of the entire expression.

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Answer 3

To differentiate the given equation (x - \cos(x^2) + \frac{y^2}{x} + 3x^5 = 4x^3) with respect to (x), we follow these steps:

  1. Differentiate each term of the equation with respect to (x), using the rules of differentiation.

  2. Simplify the derivatives obtained in step 1.

Let's differentiate each term of the equation:

  1. Differentiate (x) with respect to (x): [ \frac{d}{dx}(x) = 1 ]

  2. Differentiate (-\cos(x^2)) with respect to (x), using the chain rule: [ \frac{d}{dx}(-\cos(x^2)) = -(-\sin(x^2))(2x) = 2x\sin(x^2) ]

  3. Differentiate (\frac{y^2}{x}) with respect to (x), using the quotient rule: [ \frac{d}{dx}\left(\frac{y^2}{x}\right) = \frac{x(2y\frac{dy}{dx}) - y^2(1)}{x^2} ]

  4. Differentiate (3x^5) with respect to (x): [ \frac{d}{dx}(3x^5) = 15x^4 ]

  5. Differentiate (4x^3) with respect to (x): [ \frac{d}{dx}(4x^3) = 12x^2 ]

Now, we'll substitute the derivatives obtained into the equation and simplify:

[ 1 - 2x\sin(x^2) + \frac{x(2y\frac{dy}{dx}) - y^2}{x^2} + 15x^4 = 12x^2 ]

This equation represents the derivative of the given equation 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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