derivative of 4xy

Understand the Problem

The question is asking for the derivative of the expression 4xy. To solve this, we will apply the product rule of differentiation, which states that the derivative of a product of two functions is the first function times the derivative of the second function plus the second function times the derivative of the first function.

Answer

The derivative is given by $ \frac{d(4xy)}{dx} = 4xy' + 4y $.
Answer for screen readers

The derivative of the expression $4xy$ with respect to $x$ is given by: $$ \frac{d(4xy)}{dx} = 4xy' + 4y $$

Steps to Solve

  1. Identify the functions The expression $4xy$ is a product of two functions: $u = 4x$ and $v = y$.

  2. Apply the product rule According to the product rule, the derivative of a product $uv$ is given by: $$ \frac{d(uv)}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} $$ Here, we will find $u' = \frac{d(4x)}{dx}$ and $v' = \frac{dy}{dx}$.

  3. Differentiate the first function We differentiate $u = 4x$: $$ u' = \frac{d(4x)}{dx} = 4 $$

  4. Differentiate the second function Assuming $y$ is a function of $x$, we have: $$ v' = \frac{dy}{dx} = y' $$

  5. Combine the results Now substitute $u$, $u'$, $v$, and $v'$ back into the product rule formula: $$ \frac{d(4xy)}{dx} = u y' + v u' = (4x)(y') + (y)(4) $$

  6. Simplify the final expression Combining and simplifying gives: $$ \frac{d(4xy)}{dx} = 4xy' + 4y $$

The derivative of the expression $4xy$ with respect to $x$ is given by: $$ \frac{d(4xy)}{dx} = 4xy' + 4y $$

More Information

This final expression indicates how the value of the product $4xy$ changes with respect to $x$. The term $4y$ represents how $y$ contributes to the rate of change when $x$ changes, while $4xy'$ shows the effect of changes in $y$ itself on the product.

Tips

  • Forgetting to apply the product rule for derivatives correctly. Always remember to apply the rule to both functions involved.
  • Not recognizing that $y$ can be a function of $x$ and therefore including its derivative.
  • Confusing the constants with variables; ensure that differentiation is applied correctly to each part of the product.

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