October 22, 2024
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 […]
October 22, 2024
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 […]
October 22, 2024
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 […]
October 22, 2024
To find the derivative of $3x$, we may apply the power rule which states $\dfrac{d}{dx} x^n = nx^{n – 1}$. In this case, \(3x\) can be rewritten as \(3x^1\), where the coefficient \(3\) is constant and \(x\) has an exponent of \(n=1\). Solution Using the power rule: \[\begin{align*} \dfrac{d}{dx} (3x) &= 3 \cdot \dfrac{d}{dx} (x^1) […]
October 22, 2024
The derivative of $\cot x$ is $\boxed{-\csc^2 x}$. We will use our knowledge of the derivatives of $\sin x$ and $\cos x$ to prove this result. Recall that \begin{align*} \dfrac d{dx} \sin x = \cos x\end{align*} and \begin{align*} \dfrac d{dx} \cos x = -\sin x\end{align*} For a detailed discussion and proof of the derivative of […]
October 22, 2024
The derivative of $\csc x$ is $\boxed{-\csc x\cot x}$. We will use our knowledge of the derivatives of $\sin x$ and $\cos x$ to prove this result. Recall that \begin{align*} \dfrac d{dx} \sin x = \cos x\end{align*} and \begin{align*} \dfrac d{dx} \cos x = -\sin x\end{align*} For a detailed discussion and proof of the derivative […]
October 22, 2024
The derivative of $\sec x$ is $\boxed{\sec x\tan x}$. We will use our knowledge of the derivatives of $\sin x$ and $\cos x$ to prove this result. Recall that \begin{align*} \dfrac d{dx} \sin x = \cos x\end{align*} and \begin{align*} \dfrac d{dx} \cos x = -\sin x\end{align*} For a detailed discussion and proof of the derivative […]
October 22, 2024
The derivative of $\tan x$ is $\boxed{\sec^2 x}$. We will use our knowledge of the derivatives of $\sin x$ and $\cos x$ to prove this result. Recall that \begin{align*} \dfrac d{dx} \sin x = \cos x\end{align*} and \begin{align*} \dfrac d{dx} \cos x = -\sin x\end{align*} For a detailed discussion and proof of the derivative of […]
October 22, 2024
The derivative of \( \dfrac{x+1}{x} \) is \( \boxed{-\dfrac{1}{x^2}} \). While we can use the \textbf{quotient rule}, splitting the expression into two terms provides a simpler approach. Method 1: Simplify First (Preferred Method) Rewrite \( \dfrac{x+1}{x} \) as: \[\begin{align*}\frac{x+1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + \frac{1}{x}.\end{align*}\] Now differentiate term-by-term: \[\begin{align*}\frac{d}{dx} \left( 1 + \frac{1}{x} […]
