Answer :
To differentiate the function [tex]\( y = \sqrt{x^3} \left( x^{-2} + x \right)^2 \)[/tex], we need to apply the product rule, chain rule, and some algebraic simplifications. Here is the detailed step-by-step solution:
1. Identify the components of the function.
The function [tex]\( y \)[/tex] can be broken down into two main components:
[tex]\[ y = u \cdot v \][/tex]
where
[tex]\[ u = \sqrt{x^3} \quad \text{and} \quad v = (x^{-2} + x)^2. \][/tex]
2. Differentiate [tex]\( u \)[/tex] and [tex]\( v \)[/tex] separately.
Let's differentiate [tex]\( u = \sqrt{x^3} \)[/tex] first.
[tex]\[ u = (x^3)^{1/2} = x^{3/2} \][/tex]
Using the power rule, we have:
[tex]\[ \frac{du}{dx} = \frac{3}{2} x^{1/2}. \][/tex]
Next, differentiate [tex]\( v = (x^{-2} + x)^2 \)[/tex] using the chain rule.
Let [tex]\( w = x^{-2} + x \)[/tex], so [tex]\( v = w^2 \)[/tex].
The derivative of [tex]\( w \)[/tex] is:
[tex]\[ \frac{dw}{dx} = -2x^{-3} + 1. \][/tex]
Therefore, applying the chain rule:
[tex]\[ \frac{dv}{dx} = 2w \cdot \frac{dw}{dx} = 2(x^{-2} + x)(-2x^{-3} + 1). \][/tex]
3. Apply the product rule:
According to the product rule:
[tex]\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}. \][/tex]
Substitute [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( \frac{du}{dx} \)[/tex], and [tex]\( \frac{dv}{dx} \)[/tex] into the product rule:
[tex]\[ u = x^{3/2}, \quad v = (x^{-2} + x)^2, \][/tex]
[tex]\[ \frac{du}{dx} = \frac{3}{2}x^{1/2}, \quad \frac{dv}{dx} = 2(x^{-2} + x)(-2x^{-3} + 1). \][/tex]
[tex]\[ \frac{dy}{dx} = x^{3/2} \cdot 2(x^{-2} + x)(-2x^{-3} + 1) + (x^{-2} + x)^2 \cdot \frac{3}{2}x^{1/2}. \][/tex]
4. Simplify the expression:
Factor the terms to get a simplified form (this requires some algebraic manipulation which I've done initially):
[tex]\[ \frac{dy}{dx} = (2 - \frac{4}{x^3}) (x + \frac{1}{x^2}) \sqrt{x^3} + \frac{3}{2x} (x + \frac{1}{x^2})^2 \sqrt{x^3}. \][/tex]
This is your final derivative:
[tex]\[ \boxed{\left(2 - \frac{4}{x^3}\right) \left(x + \frac{1}{x^2}\right) \sqrt{x^3} + \frac{3}{2x} \left(x + \frac{1}{x^2}\right)^2 \sqrt{x^3}}. \][/tex]
So, the derivative of the function [tex]\( y = \sqrt{x^3} (x^{-2} + x)^2 \)[/tex] is as presented.
1. Identify the components of the function.
The function [tex]\( y \)[/tex] can be broken down into two main components:
[tex]\[ y = u \cdot v \][/tex]
where
[tex]\[ u = \sqrt{x^3} \quad \text{and} \quad v = (x^{-2} + x)^2. \][/tex]
2. Differentiate [tex]\( u \)[/tex] and [tex]\( v \)[/tex] separately.
Let's differentiate [tex]\( u = \sqrt{x^3} \)[/tex] first.
[tex]\[ u = (x^3)^{1/2} = x^{3/2} \][/tex]
Using the power rule, we have:
[tex]\[ \frac{du}{dx} = \frac{3}{2} x^{1/2}. \][/tex]
Next, differentiate [tex]\( v = (x^{-2} + x)^2 \)[/tex] using the chain rule.
Let [tex]\( w = x^{-2} + x \)[/tex], so [tex]\( v = w^2 \)[/tex].
The derivative of [tex]\( w \)[/tex] is:
[tex]\[ \frac{dw}{dx} = -2x^{-3} + 1. \][/tex]
Therefore, applying the chain rule:
[tex]\[ \frac{dv}{dx} = 2w \cdot \frac{dw}{dx} = 2(x^{-2} + x)(-2x^{-3} + 1). \][/tex]
3. Apply the product rule:
According to the product rule:
[tex]\[ \frac{dy}{dx} = u \cdot \frac{dv}{dx} + v \cdot \frac{du}{dx}. \][/tex]
Substitute [tex]\( u \)[/tex], [tex]\( v \)[/tex], [tex]\( \frac{du}{dx} \)[/tex], and [tex]\( \frac{dv}{dx} \)[/tex] into the product rule:
[tex]\[ u = x^{3/2}, \quad v = (x^{-2} + x)^2, \][/tex]
[tex]\[ \frac{du}{dx} = \frac{3}{2}x^{1/2}, \quad \frac{dv}{dx} = 2(x^{-2} + x)(-2x^{-3} + 1). \][/tex]
[tex]\[ \frac{dy}{dx} = x^{3/2} \cdot 2(x^{-2} + x)(-2x^{-3} + 1) + (x^{-2} + x)^2 \cdot \frac{3}{2}x^{1/2}. \][/tex]
4. Simplify the expression:
Factor the terms to get a simplified form (this requires some algebraic manipulation which I've done initially):
[tex]\[ \frac{dy}{dx} = (2 - \frac{4}{x^3}) (x + \frac{1}{x^2}) \sqrt{x^3} + \frac{3}{2x} (x + \frac{1}{x^2})^2 \sqrt{x^3}. \][/tex]
This is your final derivative:
[tex]\[ \boxed{\left(2 - \frac{4}{x^3}\right) \left(x + \frac{1}{x^2}\right) \sqrt{x^3} + \frac{3}{2x} \left(x + \frac{1}{x^2}\right)^2 \sqrt{x^3}}. \][/tex]
So, the derivative of the function [tex]\( y = \sqrt{x^3} (x^{-2} + x)^2 \)[/tex] is as presented.
Answer:
[tex]\dfrac{\text{d}y}{\text{d}x}=\dfrac{7x^6+2x^3-5}{2x^3\sqrt{x}}[/tex]
Step-by-step explanation:
Given equation:
[tex]y=\sqrt{x^3}\left(x^{-2} + x\right)^2[/tex]
To differentiate the given equation, we can use the product rule:
[tex]\boxed{\begin{array}{c}\underline{\textsf{Product Rule fo Differentiation}}\\\\\textsf{If}\;y=uv\;\textsf{then:}\\\\\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x}\end{array}}[/tex]
[tex]\textsf{Let} \; u=\sqrt{x^3}=x^{\frac{3}{2}}[/tex]
[tex]\textsf{Let} \; v=\left(x^{-2}+x\right)^2[/tex]
Differentiate u with respect to x using the power rule:
[tex]\dfrac{\text{d}u}{\text{d}x}=\dfrac{3}{2} \cdot x^{\frac{3}{2}-1}\\\\\\\dfrac{\text{d}u}{\text{d}x}=\dfrac{3}{2}x^{\frac12}[/tex]
Differentiate v with respect to x using the chain rule:
[tex]\dfrac{\text{d}v}{\text{d}x}=2\left(x^{-2}+x\right) \cdot \dfrac{\text{d}}{\text{d}x}\left(x^{-2}+x\right) \\\\\\ \dfrac{\text{d}v}{\text{d}x}=2\left(x^{-2}+x\right) \left(-2x^{-3}+1\right) \\\\\\ \dfrac{\text{d}v}{\text{d}x}=\left(2x^{-2}+2x\right) \left(-2x^{-3}+1\right) \\\\\\ \dfrac{\text{d}v}{\text{d}x}=-4x^{-5}+2x^{-2}-4x^{-2}+2x \\\\\\\dfrac{\text{d}v}{\text{d}x}=-4x^{-5}-2x^{-2}+2x[/tex]
Therefore:
[tex]\dfrac{\text{d}y}{\text{d}x}=u\dfrac{\text{d}v}{\text{d}x}+v\dfrac{\text{d}u}{\text{d}x} \\\\\\ \dfrac{\text{d}y}{\text{d}x}=x^{\frac{3}{2}} \cdot \left(-4x^{-5}-2x^{-2}+2x\right)+\left(x^{-2}+x\right)^2 \cdot \dfrac{3}{2}x^{\frac12}[/tex]
Simplify by using the product rule for exponents:
[tex]\dfrac{\text{d}y}{\text{d}x}=-4x^{-\frac72}-2x^{-\frac12}+2x^{\frac52}+\left(x^{-4}+2x^{-1}+x^2\right)\cdot \dfrac{3}{2}x^{\frac12} \\\\\\\dfrac{\text{d}y}{\text{d}x}=-4x^{-\frac72}-2x^{-\frac12}+2x^{\frac52}+\dfrac{3}{2}x^{-\frac72}+3x^{-\frac12}+\dfrac{3}{2}x^{\frac52}\\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{5}{2}x^{-\frac72}+x^{-\frac12}+\dfrac{7}{2}x^{\frac52}\\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{5}{2\sqrt{x^7}}+\dfrac{1}{\sqrt{x}}+\dfrac{7\sqrt{x^{5}}}{2}[/tex]
This can be simplified further, if required:
[tex]\dfrac{\text{d}y}{\text{d}x}=-\dfrac{5}{2x^3\sqrt{x}}+\dfrac{1}{\sqrt{x}}+\dfrac{7x^2\sqrt{x}}{2} \\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{5}{2x^3\sqrt{x}}+\dfrac{2x^3}{2x^3\sqrt{x}}+\dfrac{7x^2\sqrt{x}x^3\sqrt{x}}{2x^3\sqrt{x}} \\\\\\ \dfrac{\text{d}y}{\text{d}x}=-\dfrac{5}{2x^3\sqrt{x}}+\dfrac{2x^3}{2x^3\sqrt{x}}+\dfrac{7x^6}{2x^3\sqrt{x}} \\\\\\\dfrac{\text{d}y}{\text{d}x}=\dfrac{7x^6+2x^3-5}{2x^3\sqrt{x}}[/tex]
