1. Differentiate the following functions with respect to the given variable of the functions. $$ \begin{array}{ll} ext { a) } f(x)=3 x^{3}-2 \sqrt{x} & ext { b) } f(t)=\frac{5}{t^{4}}-4 t+\pi \ ext { c) } f(x)=(x+4)(2 x-1) & ext { d) } f(y)=\frac{6 y^{3}+y-3}{2 y^{2}}\end{array} $$

Question

1. Differentiate the following functions with respect to the given variable of the functions. $$ \begin{array}{ll}	ext { a) } f(x)=3 x^{3}-2 \sqrt{x} & 	ext { b) } f(t)=\frac{5}{t^{4}}-4 t+\pi \ 	ext { c) } f(x)=(x+4)(2 x-1) & 	ext { d) } f(y)=\frac{6 y^{3}+y-3}{2 y^{2}}\end{array} $$

UpStudy AI · Accepted Answer

Certainly! Let's differentiate each of the given functions step by step with respect to their respective variables.

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### **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:**
$$
f'(t) = -20 t^{-5} - 4 = -\frac{20}{t^{5}} - 4
$$

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### **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:
$$
f'(y) = 3 - \frac{1}{2y^{2}} + \frac{3}{y^{3}}
$$

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**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) $$ f'(y) = 3 - \dfrac{1}{2y^{2}} + \dfrac{3}{y^{3}} $$
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) $$ f'(y) = 3 - \frac{1}{2y^{2}} + \frac{3}{y^{3}} $$

Differentiate the following functions with respect to the given variable of the functions.

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a)

b)

c)

d)

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