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Derivative Functions

Derivative of Algebraic Function

Applying the Derivative Rules

Now that we have mastered the various properties of derivatives, it's time to apply them to different forms of algebraic functions. Whether it's a polynomial, a rational function, or one containing a radical, the key is to recognize the function's structure and choose the right 'tool' or property to differentiate it. Let's see how this strategy is applied in a few examples.

Using the Properties of Derivatives

Let's see how the properties of derivatives work in practice through a few examples.

Differentiating a Polynomial

Find the first derivative of y=4x2+x+2y = 4x^2 + x + 2.

Solution:

We can differentiate each term one by one using the power rule and the constant rule.

y=(42)x21+(11)x11+0y' = (4 \cdot 2)x^{2-1} + (1 \cdot 1)x^{1-1} + 0
y=8x1+1x0y' = 8x^1 + 1x^0
y=8x+1(since x0=1)y' = 8x + 1 \quad (\text{since } x^0=1)

Conquering Radical Forms

Find the first derivative of y=xx3y = x\sqrt{x^3}.

Solution:

There are two ways to solve this.

Method 1: Using the Product Rule

First, we convert the radical form to an exponent: y=xx3/2y = x \cdot x^{3/2}.

Let u(x)=xu(x) = x and v(x)=x3/2v(x) = x^{3/2}. Then, u(x)=1u'(x)=1 and v(x)=32x1/2v'(x) = \frac{3}{2}x^{1/2}.

y=u(x)v(x)+u(x)v(x)y' = u'(x)v(x) + u(x)v'(x)
y=(1)(x3/2)+(x)(32x1/2)y' = (1)(x^{3/2}) + (x)(\frac{3}{2}x^{1/2})
y=x3/2+32x3/2y' = x^{3/2} + \frac{3}{2}x^{3/2}
y=52x3/2y' = \frac{5}{2}x^{3/2}

Method 2: Simplifying First

We can simplify the function before differentiating it.

y=xx3/2=x1+3/2=x5/2y = x \cdot x^{3/2} = x^{1 + 3/2} = x^{5/2}
y=52x5/21=52x3/2y' = \frac{5}{2}x^{5/2 - 1} = \frac{5}{2}x^{3/2}

Both methods give the same result. Sometimes, simplifying the function first can make the differentiation process quicker.

Tackling Rational Functions

Find the first derivative of y=2(x2+2x)y = 2\left(\frac{x^2+2}{x}\right).

Solution:

We use the quotient rule. Let u(x)=x2+2u(x) = x^2+2 and v(x)=xv(x) = x. Then u(x)=2xu'(x) = 2x and v(x)=1v'(x) = 1.

y=2(u(x)v(x)u(x)v(x)[v(x)]2)y' = 2 \cdot \left(\frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}\right)
y=2((2x)(x)(x2+2)(1)x2)y' = 2 \cdot \left(\frac{(2x)(x) - (x^2+2)(1)}{x^2}\right)
y=2(2x2x22x2)y' = 2 \cdot \left(\frac{2x^2 - x^2 - 2}{x^2}\right)
y=2(x22x2)y' = 2 \cdot \left(\frac{x^2 - 2}{x^2}\right)
y=2(x2x22x2)=2(12x2)=24x2y' = 2 \cdot \left(\frac{x^2}{x^2} - \frac{2}{x^2}\right) = 2 \cdot \left(1 - \frac{2}{x^2}\right) = 2 - \frac{4}{x^2}

Exercises

  1. Find the first derivative of y=x+x22xy = \frac{\sqrt{x} + x^2}{2x}.

Answer Key

  1. For this problem, the easiest way is to simplify the function before differentiating it.

    Step 1: Split the Fraction

    We can break the fraction into two separate parts to make it easier.

    y=x2x+x22xy = \frac{\sqrt{x}}{2x} + \frac{x^2}{2x}

    Step 2: Simplify Each Term

    Convert the square root to the exponent x1/2x^{1/2} and use exponent properties to simplify each term.

    y=x1/22x1+x22x1y = \frac{x^{1/2}}{2x^1} + \frac{x^2}{2x^1}
    y=12x1/21+12x21y = \frac{1}{2}x^{1/2 - 1} + \frac{1}{2}x^{2-1}
    y=12x1/2+12xy = \frac{1}{2}x^{-1/2} + \frac{1}{2}x

    Step 3: Apply the Power Rule

    Once the function is simplified, we can directly differentiate it term by term.

    y=12(12x1/21)+12(1)y' = \frac{1}{2}\left(-\frac{1}{2}x^{-1/2 - 1}\right) + \frac{1}{2}(1)
    y=14x3/2+12y' = -\frac{1}{4}x^{-3/2} + \frac{1}{2}

    So, the first derivative is 1214x3/2\frac{1}{2} - \frac{1}{4}x^{-3/2}.