2.7 Q-1

Question Statement

Find $y^{2}$ for the following cases:

i. $y = 2 x^{5} - 3 x^{4} + 4 x^{3} + x - 2$

ii. $y = (2x + 5)^{3/2}$

iii. $y = \sqrt{x} + \frac{1}{\sqrt{x}}$

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Background and Explanation

Before solving, it’s important to understand the following concepts:

1. Differentiation: The derivative of a function gives the rate of change of the function with respect to the variable (in this case, $x$).

2. Power Rule: For a function of the form $f(x) = x^n$, the derivative is $f'(x) = n \cdot x^{n-1}$.

3. Chain Rule: Used when differentiating composite functions. If $y = f(g(x))$, then $\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$.

In each part, we will differentiate the function twice to find the second derivative $y''$.

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Solution

i. $y = 2 x^{5} - 3 x^{4} + 4 x^{3} + x - 2$

1. First derivative $y'$:

$$\frac{dy}{dx} = \frac{d}{dx}(2x^{5} - 3x^{4} + 4x^{3} + x - 2)$$

Applying the power rule:

$$y' = 10x^{4} - 12x^{3} + 12x^{2} + 1$$

2. Second derivative $y''$: Differentiate $y'$ again:

$$y'' = \frac{d}{dx}(10x^{4} - 12x^{3} + 12x^{2} + 1)$$

Applying the power rule again:

$$y'' = 40x^{3} - 36x^{2} + 24x$$

Thus, the second derivative is:

$$y'' = 40x^{3} - 36x^{2} + 24x$$

ii. $y = (2x + 5)^{3/2}$

1. First derivative $y'$: Use the chain rule:

$$y' = \frac{3}{2}(2x + 5)^{1/2} \cdot \frac{d}{dx}(2x + 5)$$

The derivative of $2x + 5$ is $2$, so:

$$y' = 3(2x + 5)^{1/2}$$

2. Second derivative $y''$: Differentiate $y'$ again using the chain rule:

$$y'' = \frac{d}{dx}\left( 3(2x + 5)^{1/2} \right)$$

Applying the chain rule:

$$y'' = \frac{3}{2}(2x + 5)^{-1/2} \cdot \frac{d}{dx}(2x + 5)$$

The derivative of $2x + 5$ is $2$, so:

$$y'' = \frac{3}{(2x + 5)^{1/2}}$$

Thus, the second derivative is:

$$y'' = \frac{3}{\sqrt{2x + 5}}$$

iii. $y = \sqrt{x} + \frac{1}{\sqrt{x}}$

1. First derivative $y'$: Differentiating both terms:

$$\frac{dy}{dx} = \frac{d}{dx}\left(x^{1/2} + x^{-1/2}\right)$$

Using the power rule:

$$y' = \frac{1}{2} x^{-1/2} - \frac{1}{2} x^{-3/2}$$

Simplifying:

$$y' = \frac{1}{2}\left(x^{-1/2} - x^{-3/2}\right)$$

2. Second derivative $y''$: Differentiating $y'$ again:

$$y'' = \frac{d}{dx}\left( \frac{1}{2} \left(x^{-1/2} - x^{-3/2}\right) \right)$$

Using the power rule:

$$y'' = \frac{1}{2} \left( \frac{1}{2} x^{-3/2} + \frac{3}{2} x^{-5/2} \right)$$

Simplifying:

$$y'' = -\frac{1}{4} x^{-3/2} + \frac{3}{4} x^{-5/2}$$

Thus, the second derivative is:

$$y'' = \frac{-x + 3}{4x^{3/2}}$$

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Key Formulas or Methods Used

• Power Rule: For a function $f(x) = x^n$, the derivative is $f'(x) = n \cdot x^{n-1}$.

• Chain Rule: For a composite function $y = f(g(x))$, the derivative is $y' = f'(g(x)) \cdot g'(x)$.

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Summary of Steps

1. i. Differentiate the function $y = 2 x^{5} - 3 x^{4} + 4 x^{3} + x - 2$ twice to get:

   • First derivative: $y' = 10x^{4} - 12x^{3} + 12x^{2} + 1$
   • Second derivative: $y'' = 40x^{3} - 36x^{2} + 24x$

2. ii. Differentiate $y = (2x + 5)^{3/2}$ using the chain rule:

   • First derivative: $y' = 3(2x + 5)^{1/2}$
   • Second derivative: $y'' = \frac{3}{\sqrt{2x + 5}}$

3. iii. Differentiate $y = \sqrt{x} + \frac{1}{\sqrt{x}}$ twice:

   • First derivative: $y' = \frac{1}{2}\left(x^{-1/2} - x^{-3/2}\right)$
   • Second derivative: $y'' = \frac{-x + 3}{4x^{3/2}}$