Problem 57 - Single Variable Calculus

Question

Find the derivative of the function. $$ y=x\left(5^{3 x}\right) $$

Step-by-Step Solution

Answer

The short version of the answer is: \(y'(x)= 5^{3x}(1 + 3x\cdot ln5 )\)

Step 1: Identify the two functions to apply the Product Rule

Let's define the two functions that we have a product of: Function 1: \(f(x) = x\) Function 2: \(g(x) = 5^{3x}\)

Step 2: Apply the Product Rule

The Product Rule states that if we have a function y(x) that is the product of two functions f(x)g(x), then the derivative of y(x) with respect to x is given by: \( y'(x) = f'(x)g(x) + f(x)g'(x) \) So, first, we'll find the derivatives of both functions, f'(x) and g'(x), individually.

Step 3: Find the derivative of f(x)

Since f(x) = x is a simple function, its derivative is straightforward: \( f'(x) = 1 \)

Step 4: Find the derivative of g(x) using the Chain Rule

The Chain Rule states that if we have a function y(x) = h(u(x)), then the derivative of y(x) with respect to x is given by: \( y'(x) = h'(u(x))\cdot u'(x) \) So, let's apply this rule to our g(x) function: Function: \(g(x) = 5^{3x}\) Inner function: \(u(x) = 3x\) Outer function: \(h(u) = 5^u\) First, find the derivative of both the inner and outer functions: u'(x) = 3 (derivative of 3x) h'(u) = 5^u \cdot ln5 (derivative of 5^u) Now, apply the Chain Rule to find g'(x): \( g'(x) = h'(u(x))\cdot u'(x) = 5^{3x} \cdot ln5 \cdot 3 \)

Step 5: Combine derivatives using the Product Rule

Now that we have f'(x) and g'(x), we can use the Product Rule to find the derivative of y(x): \( y'(x) = f'(x)g(x) + f(x)g'(x) = 1\cdot 5^{3x} + x\cdot 3\cdot 5^{3x}\cdot ln5 \) So the derivative of the given function is: \( y'(x)= 5^{3x}(1 + 3x\cdot ln5 ) \)

Key Concepts

Product Rule

The Product Rule is a crucial tool in calculus for finding the derivative of a product of two functions. If you have a function, say \(y = f(x) \cdot g(x)\), the Product Rule helps us differentiate it. According to the rule, the derivative \(y'(x)\) is given by the formula:\[y'(x) = f'(x)g(x) + f(x)g'(x)\]This means we need to:

• Find the derivative of the first function, \(f(x)\).
• Multiply it by the second function, \(g(x)\), without differentiating it yet.
• Next, find the derivative of the second function, \(g(x)\).
• Multiply that by the first function, \(f(x)\), in its original form.

The key takeaway here is to always differentiate one function while keeping the other in its original form and then sum the results. This rule simplifies finding derivatives of products of functions, which can initially seem complicated without it.

Chain Rule

The Chain Rule is a fundamental principle for finding derivatives of composite functions. It is especially helpful when dealing with nested functions, where one function is "inside" another. Consider a composite function \(y(x) = h(u(x))\), where \(h\) acts on an inside function \(u(x)\). The Chain Rule states:\[y'(x) = h'(u(x)) \cdot u'(x)\]In this approach:

• First, find \(u'(x)\), the derivative of the inner function \(u(x)\).
• Then, find \(h'(u)\), the derivative of the outer function \(h\), but treat \(u(x)\) as a constant.
• Finally, multiply these two derivatives together to get the overall derivative.

The Chain Rule is vital when functions are exponentiated, as it breaks down a complex differentiation procedure into simple, manageable steps by focusing on one function at a time.

Exponential Functions

Exponential functions are unique because they involve continuous growth or decay, characterized by the form \(a^{x}\), where \(a\) is a constant base. They play a crucial role in differentiation due to their distinct property: the derivative of an exponential function \(a^{u(x)}\) remains proportional to the function itself. Specifically,\[\frac{d}{dx}(a^{u(x)}) = a^{u(x)} \cdot u'(x) \cdot \ln(a) \]When differentiating, always remember:

• Identify \(u(x)\), the exponent that can be any differentiable function.
• Differentiate \(u(x)\) to find \(u'(x)\).
• Multiply \(a^{u(x)}\) by the derivative \(u'(x)\) and by \(\ln(a)\).

This feature of exponential functions, which closely relates to the Chain Rule, helps simplify the differentiation process by keeping the structure of the function almost intact during differentiation.