z = x3y − xy4; x = e5t, y = sec(5t); dz

To find the derivative \( dz \) for the given function \( z = x^3 y - x y^4 \) with respect to \( t \), where \( x = e^{5t} \) and \( y = \sec(5t) \), we will apply the chain rule.

### Step 1: Write down the expression for \( z \)
\[
z = x^3 y - x y^4
\]
Substitute \( x = e^{5t} \) and \( y = \sec(5t) \) into \( z \).

### Step 2: Derivatives of \( x \) and \( y \)
First, find the derivatives of \( x \) and \( y \) with respect to \( t \):
\[
\frac{dx}{dt} = \frac{d}{dt}(e^{5t}) = 5e^{5t}
\]
\[
\frac{dy}{dt} = \frac{d}{dt}(\sec(5t)) = 5\sec(5t)\tan(5t)
\]

### Step 3: Differentiate \( z \) using the product and chain rules

For the first term, \( x^3 y \):
\[
\frac{d}{dt}(x^3 y) = 3x^2 \frac{dx}{dt} y + x^3 \frac{dy}{dt}
\]
Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\[
\frac{d}{dt}(x^3 y) = 3x^2 (5e^{5t}) y + x^3 (5\sec(5t)\tan(5t))
\]

For the second term, \( x y^4 \):
\[
\frac{d}{dt}(x y^4) = \frac{dx}{dt} y^4 + x \cdot 4y^3 \frac{dy}{dt}
\]
Substitute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \):
\[
\frac{d}{dt}(x y^4) = (5e^{5t}) y^4 + x \cdot 4y^3 (5\sec(5t)\tan(5t))
\]

### Step 4: Combine the results
The total derivative is:
\[
\frac{dz}{dt} = \frac{d}{dt}(x^3 y) - \frac{d}{dt}(x y^4)
\]
Substitute the terms we found:
\[
\frac{dz}{dt} = \left[ 3x^2 (5e^{5t}) y + x^3 (5\sec(5t)\tan(5t)) \right] - \left[ (5e^{5t}) y^4 + x \cdot 4y^3 (5\sec(5t)\tan(5t)) \right]
\]

You can now plug in \( x = e^{5t} \) and \( y = \sec(5t) \) to simplify further if necessary.

Would you like to simplify this expression further or need a detailed explanation of any step?

### Related Questions:
1. How do you apply the product rule for differentiation?
2. What is the chain rule and how is it used in multivariable calculus?
3. How do you differentiate exponential functions?
4. How is the derivative of a trigonometric function calculated?
5. What are the properties of the secant and tangent functions?

**Tip:** When dealing with products of multiple variables, always remember to apply the product rule carefully and consider whether implicit differentiation is required.

z = x^3 y − xy^4; x = e^{5t}, y = sec(5t); find dz

This page explains how to differentiate z = x^3 y − xy^4 with respect to t, where x = e^{5t} and y = sec(5t), using the chain and product rules. Includes step-by-step derivations for advanced calculus students, covering exponential and trigonometric functions.

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