Answer
Here are the derivatives of the given functions:
a) \( f'(x) = 9x^{2} - \frac{1}{\sqrt{x}} \)
b) \( f'(t) = -\frac{20}{t^{5}} - 4 \)
c) \( f'(x) = 4x + 7 \)
d) \( f'(y) = 3 - \frac{1}{2y^{2}} + \frac{3}{y^{3}} \)
Solution
Certainly! Let's differentiate each of the given functions step by step with respect to their respective variables.
---
### **a) \( f(x) = 3x^{3} - 2 \sqrt{x} \)**
**Differentiate with respect to \( x \).**
**Solution:**
1. **Differentiate \( 3x^{3} \):**
\[
\frac{d}{dx} [3x^{3}] = 3 \times 3x^{2} = 9x^{2}
\]
2. **Differentiate \( -2 \sqrt{x} \):**
\[
\frac{d}{dx} [-2x^{1/2}] = -2 \times \frac{1}{2} x^{-1/2} = -x^{-1/2} = -\frac{1}{\sqrt{x}}
\]
**Combine the results:**
\[
f'(x) = 9x^{2} - \frac{1}{\sqrt{x}}
\]
---
### **b) \( f(t) = \frac{5}{t^{4}} - 4t + \pi \)**
**Differentiate with respect to \( t \).**
**Solution:**
1. **Rewrite the function for easier differentiation:**
\[
f(t) = 5t^{-4} - 4t + \pi
\]
2. **Differentiate each term:**
- \( \frac{d}{dt} [5t^{-4}] = 5 \times (-4) t^{-5} = -20 t^{-5} \)
- \( \frac{d}{dt} [-4t] = -4 \)
- \( \frac{d}{dt} [\pi] = 0 \) (since \( \pi \) is a constant)
**Combine the results:**
\[
f'(t) = -20 t^{-5} - 4 = -\frac{20}{t^{5}} - 4
\]
---
### **c) \( f(x) = (x + 4)(2x - 1) \)**
**Differentiate with respect to \( x \).**
**Solution:**
First, expand the function to simplify differentiation:
\[
f(x) = (x + 4)(2x - 1) = 2x^{2} - x + 8x - 4 = 2x^{2} + 7x - 4
\]
Now, differentiate term by term:
\[
f'(x) = \frac{d}{dx} [2x^{2}] + \frac{d}{dx} [7x] - \frac{d}{dx} [4] = 4x + 7 - 0 = 4x + 7
\]
**Alternatively, using the product rule:**
\[
f'(x) = (x + 4)'(2x - 1) + (x + 4)(2x - 1)'
\]
\[
f'(x) = (1)(2x - 1) + (x + 4)(2) = 2x - 1 + 2x + 8 = 4x + 7
\]
---
### **d) \( f(y) = \frac{6y^{3} + y - 3}{2y^{2}} \)**
**Differentiate with respect to \( y \).**
**Solution:**
First, simplify the function by dividing each term in the numerator by \( 2y^{2} \):
\[
f(y) = \frac{6y^{3}}{2y^{2}} + \frac{y}{2y^{2}} - \frac{3}{2y^{2}} = 3y + \frac{1}{2} y^{-1} - \frac{3}{2} y^{-2}
\]
Now, differentiate term by term:
\[
f'(y) = \frac{d}{dy} [3y] + \frac{d}{dy} \left[ \frac{1}{2} y^{-1} \right] - \frac{d}{dy} \left[ \frac{3}{2} y^{-2} \right]
\]
\[
f'(y) = 3 + \frac{1}{2} (-1) y^{-2} - \frac{3}{2} (-2) y^{-3}
\]
\[
f'(y) = 3 - \frac{1}{2} y^{-2} + 3 y^{-3}
\]
\[
f'(y) = 3 - \frac{1}{2y^{2}} + \frac{3}{y^{3}}
\]
**Alternatively, using the quotient rule:**
Given \( f(y) = \frac{u(y)}{v(y)} \) where \( u(y) = 6y^{3} + y - 3 \) and \( v(y) = 2y^{2} \),
\[
f'(y) = \frac{u'(y)v(y) - u(y)v'(y)}{[v(y)]^{2}}
\]
Calculate the derivatives:
\[
u'(y) = 18y^{2} + 1
\]
\[
v'(y) = 4y
\]
Now plug into the quotient rule:
\[
f'(y) = \frac{(18y^{2} + 1)(2y^{2}) - (6y^{3} + y - 3)(4y)}{4y^{4}}
\]
Simplify the numerator:
\[
= \frac{36y^{4} + 2y^{2} - 24y^{4} - 4y^{2} + 12y}{4y^{4}} = \frac{12y^{4} - 2y^{2} + 12y}{4y^{4}} = \frac{6y^{3} - y + 6}{2y^{4}}
\]
This simplifies to the same result as earlier:
\[
f'(y) = 3 - \frac{1}{2y^{2}} + \frac{3}{y^{3}}
\]
---
**Final Summary of Derivatives:**
a) \( f'(x) = 9x^{2} - \dfrac{1}{\sqrt{x}} \)
b) \( f'(t) = -\dfrac{20}{t^{5}} - 4 \)
c) \( f'(x) = 4x + 7 \)
d) \( f'(y) = 3 - \dfrac{1}{2y^{2}} + \dfrac{3}{y^{3}} \)
Reviewed and approved by the UpStudy tutoring team
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