I Aced Calculus Team

Master Exponent Laws with Ease

Introduction Exponent rules are the building blocks of simplifying complex calculations. Master them, and math becomes much simpler! These rules help simplify expressions, solve equations, and tackle real-world problems. Let’s break down these laws with clear examples, visuals, and practice problems to ensure they stick. Product Rule Add exponents when multiplying \[\begin{align*} x^m \cdot x^n…

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Integration by Parts

Integration by parts is the inverse of the derivative product rule. It is very useful when $u$-substitution and using standard integration techniques aren’t enough to handle the problem. Integration by parts states that if you have an integral of the form $\displaystyle \int u \: dv$, we may rewrite it as: \[\begin{align*} \int u \:…

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Integral Power Rule

There are many different ways to compute integrals. While all of the integration rules are important, they all have their place and can be used in different situations. When it comes to integrating polynomials, the inverse power rule is the most useful technique. The inverse power rule states: \[\begin{align*} \boxed{\int x^n \: dx = \dfrac{x^{n+1}}{n+1}…

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Understanding cos pi/4 or cos 45 degrees

Introduction The answer to this question is \[\begin{align*} \boxed{\cos \dfrac{\pi}{4} = \dfrac{\sqrt{2}}{2}} \end{align*}\] In this article, we’ll learn how to find \(\cos \dfrac{\pi}{4}\), which is the same as \(\cos 45^\circ\). We’ll approach this by using both a right triangle and the unit circle. After that, we’ll go through a practical example where knowing \(\cos \dfrac{\pi}{4}\)…

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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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