Question Statement

Find the derivative of $y = x^{3} e^{5 x}$ with respect to $x$ .

Background and Explanation

This problem involves differentiating a product of two functions: a polynomial $x^{3}$ and an exponential $e^{5 x}$ . You will need to apply the product rule combined with the chain rule (for the exponential term).

Solution

We are given the function: $y = x^{3} e^{5 x}$

To differentiate this, we recognize it as a product of two functions:

$u = x^{3}$ (the polynomial part)
$v = e^{5 x}$ (the exponential part)

Step 1: Apply the product rule $\frac{d}{d x} (uv) = \frac{d u}{d x} \cdot v + u \cdot \frac{d v}{d x}$

$\frac{d y}{d x} = \frac{d}{d x} (x^{3}) \cdot e^{5 x} + x^{3} \cdot \frac{d}{d x} (e^{5 x})$

Step 2: Differentiate $x^{3}$ using the power rule: $\frac{d}{d x} (x^{3}) = 3 x^{2}$

Step 3: Differentiate $e^{5 x}$ using the chain rule. The outer function is $e^{u}$ and the inner function is $u = 5 x$ : $\frac{d}{d x} (e^{5 x}) = e^{5 x} \cdot \frac{d}{d x} (5 x) = e^{5 x} \cdot 5 = 5 e^{5 x}$

Step 4: Substitute these derivatives back into the product rule expression: $\frac{d y}{d x} = 3 x^{2} e^{5 x} + x^{3} \cdot 5 e^{5 x}$

Step 5: Simplify the expression: $\frac{d y}{d x} = 3 x^{2} e^{5 x} + 5 x^{3} e^{5 x}$

Step 6: Factor out the common terms $x^{2} e^{5 x}$ : $\frac{d y}{d x} = x^{2} e^{5 x} (3 + 5 x)$

Key Formulas or Methods Used

Product Rule: $\frac{d}{d x} [u (x) \cdot v (x)] = u^{'} (x) \cdot v (x) + u (x) \cdot v^{'} (x)$
Chain Rule: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$ , specifically $\frac{d}{d x} [e^{u (x)}] = e^{u (x)} \cdot u^{'} (x)$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$

Summary of Steps

Identify the two functions being multiplied: $u = x^{3}$ and $v = e^{5 x}$
Find $u^{'} = 3 x^{2}$ using the power rule
Find $v^{'} = 5 e^{5 x}$ using the chain rule on the exponential
Apply the product rule: $\frac{d y}{d x} = u^{'} v + u v^{'}$
Substitute: $3 x^{2} e^{5 x} + x^{3} (5 e^{5 x})$
Factor out $x^{2} e^{5 x}$ to get the final answer: $x^{2} e^{5 x} (3 + 5 x)$

Question Statement

Find the derivative of $y = x^{3} e^{5 x}$ with respect to $x$ .

Background and Explanation

Solution

We are given the function: $y = x^{3} e^{5 x}$

To differentiate this, we recognize it as a product of two functions:

$u = x^{3}$ (the polynomial part)
$v = e^{5 x}$ (the exponential part)

Step 1: Apply the product rule $\frac{d}{d x} (uv) = \frac{d u}{d x} \cdot v + u \cdot \frac{d v}{d x}$

$\frac{d y}{d x} = \frac{d}{d x} (x^{3}) \cdot e^{5 x} + x^{3} \cdot \frac{d}{d x} (e^{5 x})$

Step 2: Differentiate $x^{3}$ using the power rule: $\frac{d}{d x} (x^{3}) = 3 x^{2}$

Step 4: Substitute these derivatives back into the product rule expression: $\frac{d y}{d x} = 3 x^{2} e^{5 x} + x^{3} \cdot 5 e^{5 x}$

Step 5: Simplify the expression: $\frac{d y}{d x} = 3 x^{2} e^{5 x} + 5 x^{3} e^{5 x}$

Step 6: Factor out the common terms $x^{2} e^{5 x}$ : $\frac{d y}{d x} = x^{2} e^{5 x} (3 + 5 x)$

Key Formulas or Methods Used

Product Rule: $\frac{d}{d x} [u (x) \cdot v (x)] = u^{'} (x) \cdot v (x) + u (x) \cdot v^{'} (x)$
Chain Rule: $\frac{d}{d x} [f (g (x))] = f^{'} (g (x)) \cdot g^{'} (x)$ , specifically $\frac{d}{d x} [e^{u (x)}] = e^{u (x)} \cdot u^{'} (x)$
Power Rule: $\frac{d}{d x} [x^{n}] = n x^{n - 1}$

Summary of Steps

Identify the two functions being multiplied: $u = x^{3}$ and $v = e^{5 x}$
Find $u^{'} = 3 x^{2}$ using the power rule
Find $v^{'} = 5 e^{5 x}$ using the chain rule on the exponential
Apply the product rule: $\frac{d y}{d x} = u^{'} v + u v^{'}$
Substitute: $3 x^{2} e^{5 x} + x^{3} (5 e^{5 x})$
Factor out $x^{2} e^{5 x}$ to get the final answer: $x^{2} e^{5 x} (3 + 5 x)$

2.7 Q-17

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps

2.7 Q-17

Question Statement

Background and Explanation

Solution

Key Formulas or Methods Used

Summary of Steps