Derivative Functions 2A1873

1. **Problem Statement:** Find the derivative of the function $$y = (x^2 + 1) \cdot \frac{1}{x}$$. 2. **Formula Used:** Use the product rule for derivatives: $$\frac{d}{dx}[u \cdot v] = u'v + uv'$$. 3. **Step-by-step Solution:** - Let $$u = x^2 + 1$$ and $$v = \frac{1}{x} = x^{-1}$$. - Compute derivatives: $$u' = 2x$$ and $$v' = -x^{-2}$$. - Apply product rule: $$y' = u'v + uv' = 2x \cdot x^{-1} + (x^2 + 1)(-x^{-2})$$. - Simplify terms: $$2x \cdot x^{-1} = 2$$, $$(x^2 + 1)(-x^{-2}) = -\frac{x^2 + 1}{x^2} = -1 - \frac{1}{x^2}$$. - Combine: $$y' = 2 - 1 - \frac{1}{x^2} = 1 - \frac{1}{x^2}$$. 4. **Final answer:** $$\boxed{y' = 1 - \frac{1}{x^2}}$$. 1. **Problem Statement:** Find the derivative of $$y = \frac{(x^2 + 3)(x - 1)}{x}$$. 2. **Formula Used:** Use the quotient rule: $$\frac{d}{dx}\left(\frac{f}{g}\right) = \frac{f'g - fg'}{g^2}$$. 3. **Step-by-step Solution:** - Let $$f = (x^2 + 3)(x - 1)$$ and $$g = x$$. - First, expand $$f$$: $$f = x^3 - x^2 + 3x - 3$$. - Compute derivatives: $$f' = 3x^2 - 2x + 3$$, $$g' = 1$$. - Apply quotient rule: $$y' = \frac{(3x^2 - 2x + 3) \cdot x - (x^3 - x^2 + 3x - 3) \cdot 1}{x^2}$$. - Simplify numerator: $$3x^3 - 2x^2 + 3x - x^3 + x^2 - 3x + 3 = 2x^3 - x^2 + 3$$. - Final derivative: $$y' = \frac{2x^3 - x^2 + 3}{x^2}$$. 4. **Final answer:** $$\boxed{y' = \frac{2x^3 - x^2 + 3}{x^2}}$$. 1. **Problem Statement:** Find the derivative of $$y = e^x \sin x$$. 2. **Formula Used:** Product rule and derivatives of exponential and sine: $$\frac{d}{dx} e^x = e^x$$, $$\frac{d}{dx} \sin x = \cos x$$. 3. **Step-by-step Solution:** - Let $$u = e^x$$ and $$v = \sin x$$. - Compute derivatives: $$u' = e^x$$, $$v' = \cos x$$. - Apply product rule: $$y' = u'v + uv' = e^x \sin x + e^x \cos x = e^x (\sin x + \cos x)$$. 4. **Final answer:** $$\boxed{y' = e^x (\sin x + \cos x)}$$. 1. **Problem Statement:** Find the derivative of $$y = e^{2x} \cos x$$. 2. **Formula Used:** Product rule, chain rule: $$\frac{d}{dx} e^{2x} = 2 e^{2x}$$, $$\frac{d}{dx} \cos x = -\sin x$$. 3. **Step-by-step Solution:** - Let $$u = e^{2x}$$ and $$v = \cos x$$. - Compute derivatives: $$u' = 2 e^{2x}$$, $$v' = -\sin x$$. - Apply product rule: $$y' = u'v + uv' = 2 e^{2x} \cos x - e^{2x} \sin x = e^{2x} (2 \cos x - \sin x)$$. 4. **Final answer:** $$\boxed{y' = e^{2x} (2 \cos x - \sin x)}$$. 1. **Problem Statement:** Find the derivative of $$y = e^{\sin x}$$. 2. **Formula Used:** Chain rule: $$\frac{d}{dx} e^{g(x)} = e^{g(x)} g'(x)$$, $$\frac{d}{dx} \sin x = \cos x$$. 3. **Step-by-step Solution:** - Let $$g(x) = \sin x$$. - Compute derivative: $$g'(x) = \cos x$$. - Apply chain rule: $$y' = e^{\sin x} \cos x$$. 4. **Final answer:** $$\boxed{y' = e^{\sin x} \cos x}$$. 1. **Problem Statement:** Find the derivative of $$y = \sin(e^x)$$. 2. **Formula Used:** Chain rule: $$\frac{d}{dx} \sin(g(x)) = \cos(g(x)) g'(x)$$, $$\frac{d}{dx} e^x = e^x$$. 3. **Step-by-step Solution:** - Let $$g(x) = e^x$$. - Compute derivative: $$g'(x) = e^x$$. - Apply chain rule: $$y' = \cos(e^x) e^x$$. 4. **Final answer:** $$\boxed{y' = e^x \cos(e^x)}$$. 1. **Problem Statement:** Find the derivative of $$y = x e^x$$. 2. **Formula Used:** Product rule: $$\frac{d}{dx} x = 1$$, $$\frac{d}{dx} e^x = e^x$$. 3. **Step-by-step Solution:** - Let $$u = x$$ and $$v = e^x$$. - Compute derivatives: $$u' = 1$$, $$v' = e^x$$. - Apply product rule: $$y' = u'v + uv' = e^x + x e^x = e^x (1 + x)$$. 4. **Final answer:** $$\boxed{y' = e^x (1 + x)}$$. 1. **Problem Statement:** Find the derivative of $$y = \tan(3x)$$. 2. **Formula Used:** Chain rule and derivative of tangent: $$\frac{d}{dx} \tan u = \sec^2 u \cdot u'$$, $$\frac{d}{dx} 3x = 3$$. 3. **Step-by-step Solution:** - Let $$u = 3x$$. - Compute derivative: $$u' = 3$$. - Apply chain rule: $$y' = \sec^2(3x) \cdot 3 = 3 \sec^2(3x)$$. 4. **Final answer:** $$\boxed{y' = 3 \sec^2(3x)}$$. 1. **Problem Statement:** Find the derivative of $$y = x \ln x$$. 2. **Formula Used:** Product rule and derivative of natural log: $$\frac{d}{dx} x = 1$$, $$\frac{d}{dx} \ln x = \frac{1}{x}$$. 3. **Step-by-step Solution:** - Let $$u = x$$ and $$v = \ln x$$. - Compute derivatives: $$u' = 1$$, $$v' = \frac{1}{x}$$. - Apply product rule: $$y' = u'v + uv' = \ln x + x \cdot \frac{1}{x} = \ln x + 1$$. 4. **Final answer:** $$\boxed{y' = \ln x + 1}$$. 1. **Problem Statement:** Find the derivative of $$y = \ln(x \sin x)$$. 2. **Formula Used:** Chain rule and derivative of natural log: $$\frac{d}{dx} \ln u = \frac{u'}{u}$$, product rule for $$u = x \sin x$$. 3. **Step-by-step Solution:** - Let $$u = x \sin x$$. - Compute $$u'$$ using product rule: $$u' = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$$. - Apply chain rule: $$y' = \frac{u'}{u} = \frac{\sin x + x \cos x}{x \sin x}$$. 4. **Final answer:** $$\boxed{y' = \frac{\sin x + x \cos x}{x \sin x}}$$. 1. **Problem Statement:** Find the derivative of $$y = \ln(\sin^2 x)$$. 2. **Formula Used:** Chain rule and derivative of natural log: $$\frac{d}{dx} \ln u = \frac{u'}{u}$$, $$u = \sin^2 x$$. 3. **Step-by-step Solution:** - Compute $$u'$$: $$u = (\sin x)^2$$, $$u' = 2 \sin x \cdot \cos x = 2 \sin x \cos x$$. - Apply chain rule: $$y' = \frac{u'}{u} = \frac{2 \sin x \cos x}{\sin^2 x} = 2 \frac{\cos x}{\sin x} = 2 \cot x$$. 4. **Final answer:** $$\boxed{y' = 2 \cot x}$$.