Math Exercises & Math Problems: Derivative of a Function - Free Printable

Problem Analysis:

The task appears to involve identifying or solving problems related to the given functions. Since no specific problem is stated in the image, I will assume the task is to differentiate each of the given functions with respect to \( x \). This is a common exercise when dealing with a list of functions.

Solution:

We will differentiate each function one by one. The functions are listed in alphabetical order from (a) to (Z).

---

#### (a) \( y = 1 - 2x \)
This is a linear function.
\[
\frac{dy}{dx} = \frac{d}{dx}(1) - \frac{d}{dx}(2x) = 0 - 2 = -2
\]

#### (b) \( y = \sin x \)
The derivative of \( \sin x \) is \( \cos x \).
\[
\frac{dy}{dx} = \cos x
\]

#### (c) \( y = e^x \)
The derivative of \( e^x \) is \( e^x \).
\[
\frac{dy}{dx} = e^x
\]

#### (d) \( y = \sqrt{x} \)
Rewrite \( \sqrt{x} \) as \( x^{1/2} \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x^{1/2}) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}
\]

#### (e) \( y = \frac{x^2}{2} \)
Differentiate term by term.
\[
\frac{dy}{dx} = \frac{1}{2} \cdot \frac{d}{dx}(x^2) = \frac{1}{2} \cdot 2x = x
\]

#### (f) \( y = \ln x \)
The derivative of \( \ln x \) is \( \frac{1}{x} \).
\[
\frac{dy}{dx} = \frac{1}{x}
\]

#### (g) \( y = 3^x \)
Use the formula for the derivative of an exponential function \( a^x \): \( \frac{d}{dx}(a^x) = a^x \ln a \).
\[
\frac{dy}{dx} = 3^x \ln 3
\]

#### (h) \( y = \cos x \)
The derivative of \( \cos x \) is \( -\sin x \).
\[
\frac{dy}{dx} = -\sin x
\]

#### (i) \( y = x^6 + 5x^4 + 2x^3 - x \)
Differentiate each term separately.
\[
\frac{dy}{dx} = \frac{d}{dx}(x^6) + \frac{d}{dx}(5x^4) + \frac{d}{dx}(2x^3) - \frac{d}{dx}(x)
\]
\[
= 6x^5 + 20x^3 + 6x^2 - 1
\]

#### (j) \( y = \ln(1 + 2x) \)
Use the chain rule. Let \( u = 1 + 2x \), so \( y = \ln u \).
\[
\frac{dy}{dx} = \frac{d}{du}(\ln u) \cdot \frac{du}{dx} = \frac{1}{u} \cdot 2 = \frac{1}{1 + 2x} \cdot 2 = \frac{2}{1 + 2x}
\]

#### (k) \( y = (1 + x)^6 \)
Use the chain rule. Let \( u = 1 + x \), so \( y = u^6 \).
\[
\frac{dy}{dx} = \frac{d}{du}(u^6) \cdot \frac{du}{dx} = 6u^5 \cdot 1 = 6(1 + x)^5
\]

#### (l) \( y = \frac{3}{x\pi} \)
Rewrite as \( y = \frac{3}{\pi} \cdot \frac{1}{x} \).
\[
\frac{dy}{dx} = \frac{3}{\pi} \cdot \frac{d}{dx}\left(\frac{1}{x}\right) = \frac{3}{\pi} \cdot \left(-\frac{1}{x^2}\right) = -\frac{3}{\pi x^2}
\]

#### (m) \( y = xe^x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x \) and \( v = e^x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x) \cdot e^x + x \cdot \frac{d}{dx}(e^x) = 1 \cdot e^x + x \cdot e^x = e^x + xe^x = e^x(1 + x)
\]

#### (n) \( y = \frac{1}{1 + x} \)
Rewrite as \( y = (1 + x)^{-1} \) and use the chain rule.
\[
\frac{dy}{dx} = \frac{d}{dx}((1 + x)^{-1}) = -1(1 + x)^{-2} \cdot 1 = -\frac{1}{(1 + x)^2}
\]

#### (o) \( y = \log_3 x \)
Use the change of base formula: \( \log_3 x = \frac{\ln x}{\ln 3} \).
\[
\frac{dy}{dx} = \frac{d}{dx}\left(\frac{\ln x}{\ln 3}\right) = \frac{1}{\ln 3} \cdot \frac{d}{dx}(\ln x) = \frac{1}{\ln 3} \cdot \frac{1}{x} = \frac{1}{x \ln 3}
\]

#### (p) \( y = \arctan x \)
The derivative of \( \arctan x \) is \( \frac{1}{1 + x^2} \).
\[
\frac{dy}{dx} = \frac{1}{1 + x^2}
\]

#### (q) \( y = x^2 e^x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x^2 \) and \( v = e^x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot e^x + x^2 \cdot \frac{d}{dx}(e^x) = 2x \cdot e^x + x^2 \cdot e^x = e^x(2x + x^2)
\]

#### (r) \( y = x^4 \ln x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x^4 \) and \( v = \ln x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x^4) \cdot \ln x + x^4 \cdot \frac{d}{dx}(\ln x) = 4x^3 \cdot \ln x + x^4 \cdot \frac{1}{x} = 4x^3 \ln x + x^3 = x^3(4 \ln x + 1)
\]

#### (s) \( y = x^2 \sin 2x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x^2 \) and \( v = \sin 2x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x^2) \cdot \sin 2x + x^2 \cdot \frac{d}{dx}(\sin 2x) = 2x \cdot \sin 2x + x^2 \cdot 2\cos 2x = 2x \sin 2x + 2x^2 \cos 2x = 2x(\sin 2x + x \cos 2x)
\]

#### (t) \( y = \ln^2 x \)
Let \( u = \ln x \), so \( y = u^2 \). Use the chain rule.
\[
\frac{dy}{dx} = \frac{d}{du}(u^2) \cdot \frac{du}{dx} = 2u \cdot \frac{1}{x} = 2 \ln x \cdot \frac{1}{x} = \frac{2 \ln x}{x}
\]

#### (u) \( y = e^x \sin x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = e^x \) and \( v = \sin x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(e^x) \cdot \sin x + e^x \cdot \frac{d}{dx}(\sin x) = e^x \cdot \sin x + e^x \cdot \cos x = e^x(\sin x + \cos x)
\]

#### (v) \( y = x^4 e^{2x} \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = x^4 \) and \( v = e^{2x} \).
\[
\frac{dy}{dx} = \frac{d}{dx}(x^4) \cdot e^{2x} + x^4 \cdot \frac{d}{dx}(e^{2x}) = 4x^3 \cdot e^{2x} + x^4 \cdot 2e^{2x} = e^{2x}(4x^3 + 2x^4) = 2x^3 e^{2x}(2 + x)
\]

#### (w) \( y = e^{-x} \cos x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = e^{-x} \) and \( v = \cos x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(e^{-x}) \cdot \cos x + e^{-x} \cdot \frac{d}{dx}(\cos x) = -e^{-x} \cdot \cos x + e^{-x} \cdot (-\sin x) = -e^{-x}(\cos x + \sin x)
\]

#### (x) \( y = \cos^3 x \)
Let \( u = \cos x \), so \( y = u^3 \). Use the chain rule.
\[
\frac{dy}{dx} = \frac{d}{du}(u^3) \cdot \frac{du}{dx} = 3u^2 \cdot (-\sin x) = 3 \cos^2 x \cdot (-\sin x) = -3 \cos^2 x \sin x
\]

#### (y) \( y = \frac{2x - 1}{x + 2} \)
Use the quotient rule: \( \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \). Here, \( u = 2x - 1 \) and \( v = x + 2 \).
\[
\frac{dy}{dx} = \frac{(2)(x + 2) - (2x - 1)(1)}{(x + 2)^2} = \frac{2x + 4 - 2x + 1}{(x + 2)^2} = \frac{5}{(x + 2)^2}
\]

#### (z) \( y = \arcsin x \)
The derivative of \( \arcsin x \) is \( \frac{1}{\sqrt{1 - x^2}} \).
\[
\frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}}
\]

#### (Z) \( y = e^{4x} \sin 3x \)
Use the product rule: \( \frac{d}{dx}(uv) = u'v + uv' \). Here, \( u = e^{4x} \) and \( v = \sin 3x \).
\[
\frac{dy}{dx} = \frac{d}{dx}(e^{4x}) \cdot \sin 3x + e^{4x} \cdot \frac{d}{dx}(\sin 3x) = 4e^{4x} \cdot \sin 3x + e^{4x} \cdot 3\cos 3x = e^{4x}(4 \sin 3x + 3 \cos 3x)
\]

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Final Answer:

\[
\boxed{
\begin{aligned}
&\text{(a)} \ -2, \quad \text{(b)} \ \cos x, \quad \text{(c)} \ e^x, \quad \text{(d)} \ \frac{1}{2\sqrt{x}}, \quad \text{(e)} \ x, \quad \text{(f)} \ \frac{1}{x}, \quad \text{(g)} \ 3^x \ln 3, \quad \text{(h)} \ -\sin x, \\
&\text{(i)} \ 6x^5 + 20x^3 + 6x^2 - 1, \quad \text{(j)} \ \frac{2}{1 + 2x}, \quad \text{(k)} \ 6(1 + x)^5, \quad \text{(l)} \ -\frac{3}{\pi x^2}, \quad \text{(m)} \ e^x(1 + x), \\
&\text{(n)} \ -\frac{1}{(1 + x)^2}, \quad \text{(o)} \ \frac{1}{x \ln 3}, \quad \text{(p)} \ \frac{1}{1 + x^2}, \quad \text{(q)} \ e^x(2x + x^2), \quad \text{(r)} \ x^3(4 \ln x + 1), \\
&\text{(s)} \ 2x(\sin 2x + x \cos 2x), \quad \text{(t)} \ \frac{2 \ln x}{x}, \quad \text{(u)} \ e^x(\sin x + \cos x), \quad \text{(v)} \ 2x^3 e^{2x}(2 + x), \\
&\text{(w)} \ -e^{-x}(\cos x + \sin x), \quad \text{(x)} \ -3 \cos^2 x \sin x, \quad \text{(y)} \ \frac{5}{(x + 2)^2}, \quad \text{(z)} \ \frac{1}{\sqrt{1 - x^2}}, \\
&\text{(Z)} \ e^{4x}(4 \sin 3x + 3 \cos 3x).
\end{aligned}
}
\]