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Last updated on September 1, 2025
We use the derivative of n^x, which is n^x ln(n), as a measuring tool for how the exponential function changes in response to a slight change in x. Derivatives help us calculate growth rates or decay in real-life situations. We will now talk about the derivative of n^x in detail.
We now understand the derivative of nx.
It is commonly represented as d/dx (nx) or (nx)', and its value is nx ln(n).
The function nx has a clearly defined derivative, indicating it is differentiable within its domain.
The key concepts are mentioned below:
Exponential Function: (nx where n is a constant).
Chain Rule: Rule for differentiating nx.
Natural Logarithm: ln(n) is the natural logarithm of the base n.
The derivative of nx can be denoted as d/dx (nx) or (nx)'.
The formula we use to differentiate nx is: d/dx (nx) = nx ln(n) (or) (nx)' = nx ln(n)
The formula applies to all x where n > 0 and n ≠ 1.
We can derive the derivative of nx using proofs.
To show this, we will use the logarithmic differentiation along with the rules of differentiation.
There are several methods we use to prove this, such as:
Using Logarithmic Differentiation
Using the Chain Rule
We will now demonstrate that the differentiation of nx results in nx ln(n) using the above-mentioned methods:
Using Logarithmic Differentiation
To find the derivative of nx using logarithmic differentiation, we consider f(x) = nx. Taking the natural log of both sides, we have ln(f(x)) = ln(nx) = x ln(n).
Differentiating both sides with respect to x, we obtain d/dx [ln(f(x))] = d/dx [x ln(n)] (1/f(x)) f'(x) = ln(n) f'(x) = f(x) ln(n) Substituting f(x) = nx, we get f'(x) = nx ln(n)
Hence, proved.
Using the Chain Rule
To prove the differentiation of nx using the chain rule, Consider f(x) = e(x ln(n))
By the chain rule, the derivative is: d/dx (e(x ln(n))) = e(x ln(n)) ·
d/dx (x ln(n)) = e(x ln(n)) ·
ln(n) Since e(x ln(n)) = nx,
we have: d/dx (nx) = nx ln(n)
Hence, proved.
When a function is differentiated several times, the derivatives obtained are referred to as higher-order derivatives.
Higher-order derivatives can be a little tricky.
To understand them better, think of a car where the speed changes (first derivative) and the rate at which the speed changes (second derivative) also changes.
Higher-order derivatives make it easier to understand functions like nx.
For the first derivative of a function, we write f′(x), which indicates how the function changes or its slope at a certain point.
The second derivative is derived from the first derivative, which is denoted using f′′(x).
Similarly, the third derivative, f′′′(x), is the result of the second derivative, and this pattern continues.
For the nth Derivative of nx, we generally use f n(x) for the nth derivative of a function f(x) which tells us the change in the rate of change. (continuing for higher-order derivatives).
When n < 0, the derivative is undefined for non-integer values of x because nx may not be real. When n = 1, the derivative of nx = 0, since 1x is constant.
Students frequently make mistakes when differentiating nx. These mistakes can be resolved by understanding the proper solutions. Here are a few common mistakes and ways to solve them:
Calculate the derivative of (n^x·ln(n))
Here, we have f(x) = nx·ln(n).
Using the product rule, f'(x) = u′v + uv′ In the given equation, u = nx and v = ln(n).
Let’s differentiate each term, u′ = d/dx (nx) = nx ln(n) v′ = d/dx (ln(n)) = 0
Substituting into the given equation, f'(x) = (nx ln(n))·0 + (nx)·(ln(n))
Let’s simplify terms to get the final answer, f'(x) = nx ln(n)
Thus, the derivative of the specified function is nx ln(n).
We find the derivative of the given function by dividing the function into two parts. The first step is finding its derivative and then combining them using the product rule to get the final result.
An investment grows at a rate represented by the function y = n^x where y represents the value over time x. If x = 2 years, measure the growth rate of the investment.
We have y = nx (growth function)...(1)
Now, we will differentiate the equation (1)
Take the derivative of nx: dy/dx = nx ln(n)
Given x = 2 (substitute this into the derivative) dy/dx = n2 ln(n)
Hence, we get the growth rate of the investment at x = 2 years as n2 ln(n).
We find the growth rate of the investment at x = 2 years as n2 ln(n), which means that at a given point, the value of the investment would increase at this rate.
Derive the second derivative of the function y = n^x.
The first step is to find the first derivative, dy/dx = nx ln(n)...(1)
Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [nx ln(n)]
Using the product rule, d²y/dx² = nx ln(n)·ln(n) + nx ln(n) d²y/dx² = nx (ln(n))² + nx ln(n)
Therefore, the second derivative of the function y = nx is nx (ln(n))² + nx ln(n).
We use the step-by-step process, where we start with the first derivative. Using the product rule, we differentiate nx ln(n). We then substitute and simplify the terms to find the final answer.
Prove: d/dx ((n^x)²) = 2n^(2x) ln(n).
Let’s start using the chain rule: Consider y = (nx)²
To differentiate, we use the chain rule: dy/dx = 2(nx)·d/dx [nx]
Since the derivative of nx is nx ln(n), dy/dx = 2(nx)·nx ln(n) dy/dx = 2n(2x) ln(n)
Hence proved.
In this step-by-step process, we used the chain rule to differentiate the equation. Then, we replace nx with its derivative. As a final step, we simplify the expression to derive the equation.
Solve: d/dx (n^x/x)
To differentiate the function, we use the quotient rule: d/dx (nx/x) = (d/dx (nx)·x - nx·d/dx(x))/x²
We will substitute d/dx (nx) = nx ln(n) and d/dx (x) = 1 (x nx ln(n) - nx·1) / x² = (x nx ln(n) - nx) / x²
Therefore, d/dx (nx/x) = (x nx ln(n) - nx) / x²
In this process, we differentiate the given function using the product rule and quotient rule. As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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