November 18, 2024

Derivative of $\ln x$ times $\ln x$

\[\begin{align*} \boxed{\dfrac{d}{dx} \ln x \cdot \ln x = \dfrac{2\ln x}{x}} \end{align*}\] Solving for the Derivative To find the derivative, we will use the Product Rule: \[\begin{align*} \frac{d}{dx} f(x) \cdot g(x) = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x) \end{align*}\] In our case both functions are the same, $f(x) = g(x) = \ln x$,…

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Second Derivative of $\ln x$

\[\begin{align*} \boxed{\dfrac{d}{dx} \dfrac{d}{dx} \ln x = -\dfrac1{x^{-2}}} \end{align*}\] Solving for the Derivative To find the second derivative of $\ln x$, we must first find the first derivative of $\ln x$. The first derivative $\ln x$ is a common derivative, $\dfrac1x$. This is a derivative that you should memorize! Now that we have the first derivative,…

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Derivative of $\ln\dfrac1x$

\[\begin{align*} \boxed{\dfrac{d}{dx} \ln\dfrac{1}{x} = -\dfrac1x} \end{align*}\] 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*}…

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Derivative of $\dfrac3x$

\[\begin{align*} \boxed{\dfrac{d}{dx} \left(\dfrac3x\right) = -\dfrac3{x^2}} \end{align*}\] 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…

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Derivative of $\sec(2x)$

\[\begin{align*} \boxed{\dfrac{d}{dx} \sec(2x) = 2 \cdot \sec \left(2x\right)\tan \left(2x\right)} \end{align*}\] 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…

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Integral of $\sec(2x)$

\[\begin{align*} \boxed{\int \sec(2x) \: dx = \frac{1}{2}\ln \left|\tan \left(2x\right)+\sec \left(2x\right)\right|+C} \end{align*}\] where \( C \) is the constant of integration. Solving for the Integral To solve many integrals that involve composite functions, we turn to $u$-substitution. It is a little easier to work with $\sec(x)$ instead so we set $u = 2x$ and $du =…

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Higher Order Differentiation

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…

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