Example Question - derivative of a function
Here are examples of questions we've helped users solve.
Derivative of a Function with Rational Expression
Given the function: \( h(x) = \frac{4x^3 - 7x + 8}{x} \) First, simplify \( h(x) \) by dividing each term in the numerator by \( x \): \( h(x) = 4x^2 - 7 + \frac{8}{x} \) Now, find the derivative \( h'(x) \): \( h'(x) = \frac{d}{dx}(4x^2) - \frac{d}{dx}(7) + \frac{d}{dx}\left(\frac{8}{x}\right) \) \( h'(x) = 8x - 0 - 8x^{-2} \) \( h'(x) = 8x - 8x^{-2} \) Expressing \( -8x^{-2} \) as \( -\frac{8}{x^2} \), we have: \( h'(x) = 8x - \frac{8}{x^2} \)
Finding the Derivative of a Function
Để giải câu hỏi này, chúng ta cần tìm đạo hàm của hàm số đã cho. Hàm số được cho là \( y = \frac{2}{x} - 3\sqrt{x} + \frac{1}{x^2} \). Ta sẽ tìm đạo hàm của từng thành phần trong hàm số: 1. Đạo hàm của \(\frac{2}{x}\) là \(-\frac{2}{x^2}\). 2. Đạo hàm của \(3\sqrt{x}\) hay \(3x^{\frac{1}{2}}\) là \(\frac{3}{2}x^{-\frac{1}{2}}\) hoặc \(\frac{3}{2\sqrt{x}}\). 3. Đạo hàm của \(\frac{1}{x^2}\) hay \(x^{-2}\) là \(-2x^{-3}\) hoặc \(-\frac{2}{x^3}\). Cộng tất cả các đạo hàm ta được, ta có đạo hàm của hàm số là: \( y' = -\frac{2}{x^2} - \frac{3}{2\sqrt{x}} - \frac{2}{x^3} \). Nhìn vào các đáp án, đáp án đúng cần phải có dạng tương tự. Đáp án C có cùng các phần tử với đạo hàm mà ta vừa tìm, vậy nên đáp án C là đáp án đúng: \( C) y' = -2x^{-2} - \frac{3}{2}x^{-\frac{1}{2}} - 2x^{-3} \).
Derivative of a Function Involving Square Roots and Chains
The function is \( f(x) = x \sqrt{1 + \sqrt{2x + 6}} \). To find \( f'(x) \), we'll need to use the product rule and the chain rule. The product rule states that if you have a function \( u(x) \cdot v(x) \), then the derivative of this function with respect to \( x \) is \( u'(x) \cdot v(x) + u(x) \cdot v'(x) \). The chain rule states that the derivative of a composite function \( u(v(x)) \) is \( u'(v(x)) \cdot v'(x) \). Let \( u(x) = x \) and \( v(x) = \sqrt{1 + \sqrt{2x + 6}} \). Then \( u'(x) = 1 \) since the derivative of \( x \) with respect to \( x \) is 1. For \( v(x) \), let \( h(x) = 1 + \sqrt{2x + 6} \), so \( v(x) = \sqrt{h(x)} \). First, find the derivative of \( h(x) \): \( h'(x) = 0 + \frac{1}{2} (2x + 6)^{-\frac{1}{2}} \cdot (2) = \frac{1}{\sqrt{2x + 6}} \). Next, we use the chain rule to find the derivative of \( v(x) \): \( v'(x) = \frac{1}{2} h(x)^{-\frac{1}{2}} \cdot h'(x) = \frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \). Now we can find \( f'(x) \) using the product rule: \( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) \) \( f'(x) = 1 \cdot \sqrt{1 + \sqrt{2x + 6}} + x \cdot \left(\frac{1}{2\sqrt{1 + \sqrt{2x + 6}}} \cdot \frac{1}{\sqrt{2x + 6}} \right) \) \( f'(x) = \sqrt{1 + \sqrt{2x + 6}} + \frac{x}{2(1 + \sqrt{2x + 6})\sqrt{2x + 6}} \). This is the derivative of the given function. Be sure to simplify where possible to get the final expression for \( f'(x) \).
Finding the Derivative of a Function Expression
The image shows two graphs, y = f(x) and y = g(x), intersecting on a Cartesian coordinate plane, and there are lines \( d_1 \) and \( d_2 \) which appear to be parallel to the x-axis at different y-values. The function \( h(x) \) is defined such that \( h(x + 1) = f(x) - g(x + 2) \). We are tasked with finding the value of \( h'(2) \). The first step to find \( h'(2) \) is to differentiate the equation for \( h(x + 1) \) with respect to \( x \). So, let's differentiate \( h(x + 1) \) using the chain rule: If \( u = x + 1 \), then \( du/dx = 1 \) and so, \[ h'(u) = \frac{dh}{du} \cdot \frac{du}{dx} = \frac{dh}{du} \] Now differentiate \( f(x) \) and \( g(x + 2) \) separately with respect to \( x \). Using the property \( \frac{d}{dx}f(x) = f'(x) \) and \( \frac{d}{dx}g(x+2) \) with the chain rule (\( v = x + 2 \), \( dv/dx = 1 \)), we get: \[ h'(u) = f'(x) - g'(v) \] Knowing that \( h'(u) = h'(x + 1) \), we can convert the differentiation point from \( x \) to \( u \) to find \( h'(2) \): \[ h'(2) = f'(1) - g'(4) \] To get the values of \( f'(1) \) and \( g'(4) \), we need to find the slopes of \( f(x) \) and \( g(x) \) at \( x = 1 \) and \( x = 4 \), respectively. Looking at the graph, \( f(x) \) passes through point A(1,0), and since \( d_1 \) and \( d_2 \) are parallel to the x-axis, it means the slope is 0 at \( x = 1 \) for \( f(x) \). Therefore, the slope \( f'(1) = 0 \). The graph doesn't give us \( g(4) \) directly, but we can use the fact that the line through the point B(-1, 2) is at a 45-degree angle to infer that this line, representing \( g(x) \), has a slope of 1, as a 45-degree line in the first quadrant will have a slope of 1. Hence, \( g'(x) = 1 \) for all \( x \) in the domain of \( g \), and specifically \( g'(4) = 1 \). Now, substitute the slope values into the equation for \( h'(2) \): \[ h'(2) = 0 - 1 = -1 \] Thus, the value of \( h'(2) \) is \(-1\).
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com