Problem 2 Differentiate. $$ y=x^{4} \c... [FREE SOLUTION] (2024)

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Chapter 2: Problem 2

Differentiate. $$ y=x^{4} \cdot x^{9} \text { , two ways } $$

Short Answer

Expert verified

The derivative is \( 13 x^{12} \).

Step by step solution

01

Simplify the Expression

Combine the exponents of the same base before differentiating. Given the expression: \( y = x^4 \times x^9 \), simplify it using the property of exponents \( x^a \times x^b = x^{a+b} \). Therefore, \( y = x^{4+9} = x^{13} \).

02

Apply the Power Rule

Differentiate the simplified expression \( y = x^{13} \) using the power rule, which states \( \frac{d}{dx} x^n = n x^{n-1} \). So, \( \frac{dy}{dx} = 13 x^{12} \).

03

Confirm Method 1 Result

The result using the first method is \( \frac{dy}{dx} = 13 x^{12} \).

04

Use the Product Rule

Now apply the product rule to differentiate the original expression without simplifying first. The product rule states \( \frac{d}{dx} [u \times v] = u' \times v + u \times v' \), where \( u = x^4 \) and \( v = x^9 \).

05

Differentiate Each Term

Differentiate \( u = x^4 \) to get \( u' = 4 x^3 \) and differentiate \( v = x^9 \) to get \( v' = 9 x^8 \).

06

Apply the Product Rule

Substitute these derivatives back into the product rule: \( \frac{dy}{dx} = (4 x^3) \times (x^9) + (x^4) \times (9 x^8) \).

07

Simplify the Result

Simplify the expression: \( \frac{dy}{dx} = 4 x^{3+9} + 9 x^{4+8} = 4 x^{12} + 9 x^{12} = 13 x^{12} \).

08

Confirm Method 2 Result

The result using the second method is \( \frac{dy}{dx} = 13 x^{12} \). Both methods yield the same derivative.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Power Rule

One of the most fundamental rules in differentiation is the power rule. It is used when you need to find the derivative of a function in the form of \(x^n\), where \(n\) is any real number.
The power rule states: \( \frac{d}{dx} {x^n} = n x^{n-1} \).
This means that you take the exponent, bring it in front as a coefficient, and then subtract one from the original exponent. For example, to differentiate \(x^5\), you apply the power rule like this: \( \frac{d}{dx} {x^5} = 5 x^{4} \).
This rule is straightforward and very powerful.
In our exercise, we simplified \(x^4 \times x^9\) to \(x^{13}\) and then applied the power rule to get \(13 x^{12}\).

Using the Product Rule

While the power rule is handy, sometimes expressions require a different approach, especially when you have a product of two functions.
This is where the product rule helps. The product rule states: \( \frac{d}{dx} [u \times v] = u' \times v + u \times v' \), where \(u\) and \(v\) are functions of \(x\).
In simple terms, differentiate the first function and multiply it by the second function. Then, differentiate the second function and multiply it by the first function.
In our exercise, we let \(u = x^4\) and \(v = x^9\). We found their derivatives: \(u' = 4 x^3\) and \(v' = 9 x^8\). Plugging these into the product rule, we got: \( (4 x^3) \times (x^9) + (x^4) \times (9 x^8) \) = \( 4 x^{12} + 9 x^{12} = 13 x^{12} \)
This method confirms we get the same result as simplifying first.

Simplifying Expressions Before Differentiation

Simplifying expressions can make differentiation easier.
The exercise showed how combining exponents of the same base can simplify the problem.
Given \(y = x^4 \times x^9\), we used the rule \(x^a \times x^b = x^{a+b}\) to combine the exponents: \(4 + 9 = 13\). So, \(y = x^{13}\).
This simplification helps to reduce complexity and avoid lengthy calculations.
If an expression involves simple terms that can be combined, always simplify first. This often makes applying rules like the power rule more straightforward.

Understanding Derivatives

Derivatives measure how a function changes as its input changes.
If you imagine a curve on a graph, the derivative at any point gives the slope of the tangent to the curve at that point.
Mathematically, it is the rate of change of a function. Whether you're dealing with simple powers or complex products, finding the derivative tells you how the function's value is evolving.
In our exercise, finding the derivative of \(y = x^{13}\) gave us \(13 x^{12}\). This means for very small changes in \(x\), the function \(y\) changes by roughly \(13 x^{12}\).
Understanding this helps in various fields such as physics for speed and acceleration or economics for cost and profit changes.

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Problem 2 Differentiate. $$ y=x^{4} \c... [FREE SOLUTION] (3)

Most popular questions from this chapter

Differentiate. $$ g(t)=\frac{t^{3}-1}{t^{3}+1} \csc t $$Differentiate. $$ y=\sqrt{\cos 4 t} $$Find an equation for the tangent line to the graph of \(y=\sqrt{x^{2}+3 x}\) atthe point \((1,2)\).Find \(\frac{d y}{d u}, \frac{d u}{d x}\), and \(\frac{d y}{d x}\). $$ y=\sqrt{u} \text { and } u=x^{2}-1 $$Find an equation for the tangent line to the graph of \(f(x)=\sin ^{2} x\) atthe point \((-\pi / 6,1 / 4)\)
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Problem 2 Differentiate.  
$$  
y=x^{4} \c... [FREE SOLUTION] (2024)

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