Differentiate the function using basic differentiation formulas.
(a) f(x) = π
2
(b) f(x) = 3
4
x
8
(c) f(t) = 1.4t
5 − 2.5t
2 + 6.7
(d) y = sin(t) + π cos(t)
(e) B(y) = cy−6
(f) y = x
5/3 − x
2/3
(g) y =
√
x(x − 1)
(h) S(R) = 4πR2
(i) g(u) = √
2u +
√
3u
(j) y =
√
x+x
x2
(k) f(θ) = sin(θ)
2 +
c
θ
(l) f(t) = √3
t
2 + 2√
t
3

Let's differentiate each function using basic differentiation rules.

### (a) \( f(x) = \pi \)
Since \( \pi \) is a constant, its derivative is 0.
\[
f'(x) = 0
\]

---

### (b) \( f(x) = \frac{3}{4} x^8 \)
Using the power rule, \( \frac{d}{dx}(x^n) = nx^{n-1} \):
\[
f'(x) = \frac{3}{4} \cdot 8x^{8-1} = 6x^7
\]

---

### (c) \( f(t) = 1.4t^5 - 2.5t^2 + 6.7 \)
Differentiate each term separately:
\[
f'(t) = 1.4 \cdot 5t^{5-1} - 2.5 \cdot 2t^{2-1} + 0 = 7t^4 - 5t
\]

---

### (d) \( y = \sin(t) + \pi \cos(t) \)
Differentiate each term:
\[
y' = \cos(t) - \pi \sin(t)
\]

---

### (e) \( B(y) = cy^{-6} \)
Using the power rule and constant multiple rule, where \( c \) is a constant:
\[
B'(y) = c \cdot (-6)y^{-7} = -6cy^{-7}
\]

---

### (f) \( y = x^{5/3} - x^{2/3} \)
Differentiate each term using the power rule:
\[
y' = \frac{5}{3}x^{\frac{5}{3} - 1} - \frac{2}{3}x^{\frac{2}{3} - 1} = \frac{5}{3}x^{2/3} - \frac{2}{3}x^{-1/3}
\]

---

### (g) \( y = \sqrt{x}(x - 1) \)
Rewrite the function as \( y = x^{1/2}(x - 1) \) and apply the product rule, \( (uv)' = u'v + uv' \):
\[
y' = \frac{1}{2}x^{-\frac{1}{2}}(x - 1) + x^{\frac{1}{2}} \cdot 1
\]
Simplify:
\[
y' = \frac{x - 1}{2\sqrt{x}} + \sqrt{x}
\]

---

### (h) \( S(R) = 4\pi R^2 \)
Differentiate with respect to \( R \):
\[
S'(R) = 4\pi \cdot 2R = 8\pi R
\]

---

### (i) \( g(u) = \sqrt{2u} + \sqrt{3u} \)
Rewrite as \( g(u) = (2u)^{1/2} + (3u)^{1/2} \), then differentiate:
\[
g'(u) = \frac{1}{2}(2u)^{-1/2} \cdot 2 + \frac{1}{2}(3u)^{-1/2} \cdot 3
\]
\[
= \frac{1}{\sqrt{2u}} + \frac{3}{2\sqrt{3u}}
\]

---

### (j) \( y = \frac{\sqrt{x} + x}{x^2} \)
Rewrite as \( y = \frac{x^{1/2}}{x^2} + \frac{x}{x^2} = x^{-3/2} + x^{-1} \) and differentiate:
\[
y' = -\frac{3}{2}x^{-5/2} - x^{-2} = -\frac{3}{2x^{5/2}} - \frac{1}{x^2}
\]

---

### (k) \( f(\theta) = \sin^2(\theta) + \frac{c}{\theta} \)
Differentiate \( \sin^2(\theta) \) using the chain rule and apply the power rule to \( \frac{c}{\theta} \):
\[
f'(\theta) = 2\sin(\theta)\cos(\theta) - \frac{c}{\theta^2} = \sin(2\theta) - \frac{c}{\theta^2}
\]

---

### (l) \( f(t) = \sqrt{3}t^2 + 2\sqrt{t^3} \)
Rewrite \( 2\sqrt{t^3} \) as \( 2t^{3/2} \):
\[
f'(t) = \sqrt{3} \cdot 2t + 2 \cdot \frac{3}{2}t^{1/2} = 2\sqrt{3}t + 3t^{1/2}
\]

---

Would you like further explanations on any of these steps?

### Related Questions
1. How would you apply the product rule for more complex products?
2. What are the differences between using the chain rule and the power rule?
3. How do you approach differentiating trigonometric functions?
4. Can you explain implicit differentiation with an example?
5. How would you find the second derivative of one of these functions?

### Tip
When differentiating, it’s often helpful to rewrite the function in a simpler form before applying differentiation rules.

Differentiate the function using basic differentiation formulas.
(a) f(x) = π
(b) f(x) = 3/4 x^8
(c) f(t) = 1.4t^5 − 2.5t^2 + 6.7
(d) y = sin(t) + π cos(t)
(e) B(y) = cy^−6
(f) y = x^(5/3) − x^(2/3)
(g) y = √x(x − 1)
(h) S(R) = 4πR^2
(i) g(u) = √(2u) + √(3u)
(j) y = √x + x / x^2
(k) f(θ) = sin^2(θ) + c / θ
(l) f(t) = √3 t^2 + 2√t^3

Step-by-step differentiation of various functions using basic rules, including constant functions, power functions, and trigonometric functions. This guide covers the application of the power rule, product rule, and chain rule, ideal for high school calculus or introductory university calculus.

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