November 18, 2024
\[\begin{align*} \boxed{\dfrac{d}{dx} \ln\dfrac{1}{x} = -\dfrac1x} \end{align*}\] Solving for the Derivative Notice that $\ln\dfrac1x = -\ln x$. We can see this if we look at $\ln 1 = 0$. We may rewrite the inside: \[\begin{align*} \ln 1 &= \ln\left(\dfrac{x}{x}\right)\\ &= \ln\left(x \cdot \dfrac1x\right)\\ \end{align*}\] We may then apply the product rule for logs to obtain: \[\begin{align*} […]
November 18, 2024
\[\begin{align*} \boxed{\dfrac{d}{dx} \left(\dfrac3x\right) = -\dfrac3{x^2}} \end{align*}\] Solving for the Derivative We may pull the constant, $3$, out of the derivative and focus on $\dfrac{d}{dx} \left(\dfrac1x\right)$. This is the same as $x^{-1}$, so we may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n-1} \end{align*}\] Doing this with $x^{-1}$, we get our derivative is equal to […]
November 18, 2024
\[\begin{align*} \boxed{\dfrac{d}{dx} \sec(2x) = 2 \cdot \sec \left(2x\right)\tan \left(2x\right)} \end{align*}\] Solving for the Derivative Applying the chain rule: \[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x)) \end{align*}\] We may set our $g(x) = 2x$ and our $f(x) = \sec x$. Doing this, we get that \[\begin{align*} \dfrac{d}{dx} \sec(2x) &= \sec \left(2x\right)\tan \left(2x\right)\frac{d}{dx}\left(2x\right)\\ &= \boxed{2 \cdot \sec \left(2x\right)\tan […]
November 18, 2024
Introduction In calculus, taking the derivative of a function allows us to understand the rate of change of a function. This process allows us to gain information about the nature of a function, giving us a variety of applications to different problems. When we want to take the derivative of a function multiple times, we […]
November 15, 2024
What is Implicit Differentiation? In many calculus problems, you’ll see equations that don’t exactly represent functions. Normally, we’re used to seeing equations like $$y = 2x + 1$$ where $y$ is isolated on one side. However, what if \(x\) and \(y\) are mixed together, like $$x^2 + y^2 = 1 { or } 3x^2y -4x\cos […]
November 8, 2024
\[\begin{align*} \boxed{\frac{d}{dx} \left( 2x \ln(x) \right) = 2 \ln(x) + 2} \end{align*}\] Introduction To find the derivative of \( 2x \ln(x) \), we will use the product rule. Learning to work with logarithmic expressions is important in calculus as they show up later when tackling integration. Step-by-Step Solution The product rule, which states that if […]
November 7, 2024
Answer \[\begin{align*} \boxed{\frac{d}{dx}(\ln x) = \frac{1}{x}} \end{align*}\] The Derivative of \(\ln x\) The function $\ln x$ is called the natural logarithm function. More specifically, $\ln x = \ln_e x$, where $e \approx 2.718$. The derivative of $\ln x$ is: \[\begin{align*} \frac{d}{dx}(\ln x) = \frac{1}{x} \end{align*}\] However, some sources also use $\log x$ to describe $\ln […]
November 7, 2024
Introduction In calculus, an exponential function refers to any function that is a base with an exponent of some expression of $x$. Examples include $f(x) = 2^x$, $3^x$, $e^{x – 1}$, $5^{\sqrt{x + 1}}$, and so on. Derivatives of exponential functions often appear in calculus. The Key Idea The derivative of an exponential function $f(x) […]
November 7, 2024
In this article, we are going to prove that the derivative of $e^x$ is equal to $\boxed{e^x}$. To find the derivative of $e^x$, we must first define $e$. The formal definition of $e$ is: \[\begin{align*} e = \lim_{n\to\infty}(1+\dfrac1n)^n \end{align*}\] $e$ is approximately equal to $2.71828$. This value does not need to me memorized, as it […]
