Derivatives - Page 3

Derivative of $\dfrac{x}{5}$

The derivative of $\dfrac{x}{5}$ is $\boxed{\dfrac{1}{5}}$. To find the derivative of $\dfrac{x}{5}$, we may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n – 1} \end{align*}\] In this case, $\dfrac{x}{5}$ can be rewritten as $\dfrac{1}{5} \cdot x^1$, where the coefficient $\dfrac15$ is constant and \(x\) has an exponent of \(n=1\). Solution Using the power rule:…

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Derivative of x over 3

The derivative of $\dfrac{x}{3}$ is $\boxed{\dfrac{1}{3}}$. To find the derivative of $\dfrac{x}{3}$, we may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n – 1} \end{align*}\] In this case, $\dfrac{x}{3}$ can be rewritten as $\dfrac{1}{3} \cdot x^1$, where the coefficient $\dfrac13$ is constant and \(x\) has an exponent of \(n=1\). Solution Using the power rule:…

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Derivative of 5x

The derivative of $5x$ is $\boxed{5}$. To find the derivative of $5x$, we may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n – 1} \end{align*}\] In this case, \(5x\) can be rewritten as \(5x^1\), where the coefficient \(5\) is constant and \(x\) has an exponent of \(n=1\). Solution Using the power rule: \[\begin{align*} \dfrac{d}{dx}…

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Derivative of 4x

The derivative of $4x$ is $\boxed{4}$. To find the derivative of $4x$, we may apply the power rule: \[\begin{align*} \dfrac{d}{dx} x^n = nx^{n – 1} \end{align*}\] In this case, \(4x\) can be rewritten as \(4x^1\), where the coefficient \(4\) is constant and \(x\) has an exponent of \(n=1\). Solution Using the power rule: \[\begin{align*} \dfrac{d}{dx}…

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Derivative of $\dfrac1{1 + \sin x}$

The derivative of $\dfrac1{1 + \sin x}$ is $\boxed{-\dfrac{\cos x}{(1 + \sin x)^2}}$. To find the derivative of $\dfrac1{1 + \sin x}$, we will use a combination of the power rule and chain rule. The power rule is: \[\begin{align*} \dfrac{d}{dx} x^n &= nx^{n – 1} \end{align*}\] The chain rule is: \[\begin{align*} \dfrac{d}{dx}(f(g(x))) = f'(g(x)) \cdot…

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Derivative of 1/x^2

The derivative of $\dfrac1{x^2}$ is $\boxed{-\dfrac2{x^3}}$. To find the derivative of $\dfrac1{x^2}$, we will use the Power Rule: \[\begin{align*} \dfrac{d}{dx} x^n &= nx^{n – 1} \end{align*}\] Notice that $\dfrac1{x^2}$ may be rewritten as $x^{-2}$, making the use of the power rule quite straightforward. \[\begin{align*} \dfrac{d}{dx} \dfrac1{x^2} &= \dfrac{d}{dx} x^{-2}\\ &= -2x^{-2 – 1}\\ &= -2x^{-3}\\…

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Derivative Notations

There are a couple of notations when it comes to denoting the derivative of functions. The most common notations you will see include: $\bullet$ Leibniz Notation: $\dfrac{d}{dx} f(x)$ $\bullet$ Lagrange Notation: $f'(x)$ $\bullet$ Euler Notation: $D f(x)$ $\bullet$ Newton Notation: $\dot y$ Most likely, you will find Leibniz and Lagrange’s notations being used interchangeably in…

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Derivative of $\dfrac1{1 – x}$

To find the derivative of $\dfrac{1}{1-x}$, we will apply the chain rule $$\dfrac{d}{dx}(f(g(x)) = f'(g(x))g'(x)$$ on the functions $f(x) = \dfrac1x$ and $g(x) = 1-x$. First, we need to find the derivatives of $f(x)$ and $g(x)$: $$f'(x) = \dfrac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}$$ and $$g'(x) = \dfrac{d}{dx}(1-x) = -1$$ We are now ready to apply…

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Derivative of $\dfrac1{x – 1}$

To find the derivative of $\dfrac{1}{x – 1}$, we will apply the chain rule $$\dfrac{d}{dx}(f(g(x)) = f'(g(x))g'(x)$$ on the functions $f(x) = \dfrac1x$ and $g(x) = x – 1$. First, we need to find the derivatives of $f(x)$ and $g(x)$: $$f'(x) = \dfrac{d}{dx}(x^{-1}) = -x^{-2} = -\frac{1}{x^2}$$ and $$g'(x) = \dfrac{d}{dx}(x-1) = 1$$ We are…

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Derivative of 1/(x + 2)

Using the Quotient Rule One way to find the derivative of $\dfrac{1}{x + 2}$ is by using the quotient rule. Indeed, when you see a quotient, it is only natural to use the quotient rule. The quotient rule states that for functions $f(x)$ and $g(x)$, $$\dfrac d{dx}\left(\frac {f(x)}{g(x)}\right) = \frac{f'(x)g(x) – f(x)g'(x)}{g^2(x)}$$ If we let…

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