Derivatives

Derivative of ln(x^3)

  We will be trying to find the derivative of $\ln(x^3)$. Using the Power Rule for logarithms, which tells us $\ln(x^n) = n \ln (x)$, we get $\ln(x^3) = 3\ln(x)$. Using the fact that $\dfrac{d}{dx} (\ln(x)) = \dfrac{1}{x}$, we have \[\begin{align*} \dfrac{d}{dx}(\ln(x^3)) = \dfrac{d}{dx}(3\ln(x)) = 3 \cdot \dfrac{d}{dx} (\ln (x)) = 3 \cdot \dfrac{1}{x} =…

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Derivative of ln(x^2)

We will be trying to find the derivative of $\ln(x^2)$. Recall the formula for the Chain Rule, which states that \[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot g'(x). \end{align*}\] Using this formula, we let $f(x) = \ln(x)$ and $g(x) = x^2$ so that $f(g(x)) = \ln(g(x)) = \ln(x^2)$. To plug these values into the chain rule, we…

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Derivative of a^x

We will be finding the derivative of $a^x$ where $a$ is a constant. We will start by rewriting $a^x$ as: \[\begin{align*} a^x = (e^{\ln a})^x = e^{\ln a \cdot x}. \end{align*}\] Now, recall the formula $\dfrac{d}{dx}(e^{nx}) = e^{nx} \cdot n$. In the context of our problem, $n = \ln a$. More information on this formula…

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Derivative of -sin x

We will be finding the derivative of $-\sin x$. We first rewrite this expression as: \[\begin{align*} -\sin x = -1 \cdot \sin x. \end{align*}\] We can remove the $-1$ from the expression because it is a constant. Then, we recall that: \[\begin{align*} \dfrac{d}{dx}(\sin x) = \cos x. \end{align*}\] To learn about how we can prove…

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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 should review the product rule: \[\begin{align*} \frac{d}{dx} f(x) \cdot g(x) = f(x) \cdot \frac{d}{dx} g(x) + g(x) \cdot \frac{d}{dx} f(x) \end{align*}\] Proceeding with our two functions, $f(x) = g(x) = \ln x$, we get that…

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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 x over 3

\[\begin{align*} \boxed{\dfrac{d}{dx} \left(\dfrac{x}{3}\right) = \dfrac13} \end{align*}\] Solving for the Derivative We may pull the constant, $\dfrac13$, out of the derivative and focus on $\dfrac{d}{dx} x$. We may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n-1} \end{align*}\] Doing this with $x$, which is the same as $x^1$, we get our derivative is equal to \[\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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