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Différenciation implicit function containing product and quotient. d/dx (2y/5x)

User Asprin
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Answer:

To differentiate the given expression, we will need to use both the product rule and the quotient rule.

Let's first rewrite the expression using the reciprocal identity for x:

2y/5x = 2y * (5x)^(-1)

Now we can use the product rule and the quotient rule to find the derivative with respect to x:

d/dx [2y * (5x)^(-1)] = 2y * d/dx[(5x)^(-1)] + (5x)^(-1) * d/dx[2y]

Using the chain rule, we can find the derivative of (5x)^(-1) with respect to x:

d/dx[(5x)^(-1)] = -1/(5x)^2 * 5 * (dx/dx)

d/dx[(5x)^(-1)] = -1/(5x)^2

Using the chain rule again, we can find the derivative of 2y with respect to x:

d/dx[2y] = 2 * (dy/dx)

Substituting these expressions back into the original equation, we get:

d/dx [2y/5x] = 2y * (-1/(5x)^2) + (5x)^(-1) * 2 * (dy/dx)

Simplifying this expression, we get:

d/dx [2y/5x] = -2y/(5x)^2 + 2/(5x) * (dy/dx)

Therefore, the derivative of 2y/5x with respect to x is -2y/(5x)^2 + 2/(5x) * (dy/dx).

User Pius
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