Solved MAT1475 Calculus I Derivative worksheet Differentiate ... - Free Printable

Problem: Differentiate the following functions. Show all your steps. Answer must be simplified.

We will solve a few of these problems step by step to demonstrate the process. Let's start with the first few functions:

---

#### 1. \( f(x) = 3x^5 - 5x^3 \)

To differentiate this function, we use the power rule:
\[
\frac{d}{dx} [x^n] = nx^{n-1}
\]

Differentiate term by term:
\[
f'(x) = \frac{d}{dx} [3x^5] - \frac{d}{dx} [5x^3]
\]

Using the power rule:
\[
\frac{d}{dx} [3x^5] = 3 \cdot 5x^{5-1} = 15x^4
\]
\[
\frac{d}{dx} [5x^3] = 5 \cdot 3x^{3-1} = 15x^2
\]

Thus:
\[
f'(x) = 15x^4 - 15x^2
\]

Simplify:
\[
f'(x) = 15x^2(x^2 - 1)
\]

Final Answer:
\[
\boxed{15x^2(x^2 - 1)}
\]

---

#### 2. \( f(x) = 2x^2 (x^3 + 2)^4 \)

This function involves both the product rule and the chain rule. The product rule states:
\[
\frac{d}{dx} [u(x)v(x)] = u'(x)v(x) + u(x)v'(x)
\]

Let:
\[
u(x) = 2x^2 \quad \text{and} \quad v(x) = (x^3 + 2)^4
\]

First, differentiate \( u(x) \):
\[
u'(x) = \frac{d}{dx} [2x^2] = 2 \cdot 2x = 4x
\]

Next, differentiate \( v(x) \) using the chain rule. Let \( w(x) = x^3 + 2 \), so \( v(x) = w(x)^4 \). Then:
\[
v'(x) = \frac{d}{dx} [w(x)^4] = 4w(x)^3 \cdot w'(x)
\]
\[
w'(x) = \frac{d}{dx} [x^3 + 2] = 3x^2
\]

Thus:
\[
v'(x) = 4(x^3 + 2)^3 \cdot 3x^2 = 12x^2(x^3 + 2)^3
\]

Now apply the product rule:
\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]
\[
f'(x) = (4x)(x^3 + 2)^4 + (2x^2)(12x^2(x^3 + 2)^3)
\]

Simplify each term:
\[
f'(x) = 4x(x^3 + 2)^4 + 24x^4(x^3 + 2)^3
\]

Factor out the common term \( 4x(x^3 + 2)^3 \):
\[
f'(x) = 4x(x^3 + 2)^3 \left[ (x^3 + 2) + 6x^3 \right]
\]
\[
f'(x) = 4x(x^3 + 2)^3 \left[ x^3 + 2 + 6x^3 \right]
\]
\[
f'(x) = 4x(x^3 + 2)^3 \left[ 7x^3 + 2 \right]
\]

Final Answer:
\[
\boxed{4x(x^3 + 2)^3 (7x^3 + 2)}
\]

---

#### 3. \( f(x) = e^x \sin(2x) \)

This function involves the product rule and the chain rule. Let:
\[
u(x) = e^x \quad \text{and} \quad v(x) = \sin(2x)
\]

First, differentiate \( u(x) \):
\[
u'(x) = \frac{d}{dx} [e^x] = e^x
\]

Next, differentiate \( v(x) \) using the chain rule. Let \( w(x) = 2x \), so \( v(x) = \sin(w(x)) \). Then:
\[
v'(x) = \frac{d}{dx} [\sin(w(x))] = \cos(w(x)) \cdot w'(x)
\]
\[
w'(x) = \frac{d}{dx} [2x] = 2
\]

Thus:
\[
v'(x) = \cos(2x) \cdot 2 = 2\cos(2x)
\]

Now apply the product rule:
\[
f'(x) = u'(x)v(x) + u(x)v'(x)
\]
\[
f'(x) = (e^x)(\sin(2x)) + (e^x)(2\cos(2x))
\]

Simplify:
\[
f'(x) = e^x \sin(2x) + 2e^x \cos(2x)
\]

Factor out \( e^x \):
\[
f'(x) = e^x (\sin(2x) + 2\cos(2x))
\]

Final Answer:
\[
\boxed{e^x (\sin(2x) + 2\cos(2x))}
\]

---

#### 4. \( f(x) = (4x^3 - 2x)^{-2} \)

This function involves the chain rule. Let:
\[
u(x) = 4x^3 - 2x
\]

Then:
\[
f(x) = u(x)^{-2}
\]

Differentiate using the chain rule:
\[
f'(x) = \frac{d}{dx} [u(x)^{-2}] = -2u(x)^{-3} \cdot u'(x)
\]

First, find \( u'(x) \):
\[
u'(x) = \frac{d}{dx} [4x^3 - 2x] = 12x^2 - 2
\]

Substitute back:
\[
f'(x) = -2(4x^3 - 2x)^{-3} \cdot (12x^2 - 2)
\]

Simplify:
\[
f'(x) = -2(12x^2 - 2)(4x^3 - 2x)^{-3}
\]

Write in a more standard form:
\[
f'(x) = \frac{-2(12x^2 - 2)}{(4x^3 - 2x)^3}
\]

Factor out 2 from the numerator:
\[
f'(x) = \frac{-2 \cdot 2(6x^2 - 1)}{(4x^3 - 2x)^3}
\]
\[
f'(x) = \frac{-4(6x^2 - 1)}{(4x^3 - 2x)^3}
\]

Final Answer:
\[
\boxed{\frac{-4(6x^2 - 1)}{(4x^3 - 2x)^3}}
\]

---

Summary of Solutions:

1. \( f'(x) = 15x^2(x^2 - 1) \)
2. \( f'(x) = 4x(x^3 + 2)^3 (7x^3 + 2) \)
3. \( f'(x) = e^x (\sin(2x) + 2\cos(2x)) \)
4. \( f'(x) = \frac{-4(6x^2 - 1)}{(4x^3 - 2x)^3} \)

If you need solutions for more functions, please let me know!