Example Question - second derivative
Here are examples of questions we've helped users solve.
Partial Derivative Operations of Logarithmic Function
题目要求对函数 \( z = \ln(xy) \) 进行偏导运算。 首先，找到 \( \frac{\partial^3 z}{\partial x^2 \partial y} \) 和 \( \frac{\partial^3 z}{\partial x \partial y^2} \)，然后计算它们的差值。 函数 \( z \) 关于 \( x \) 的一阶偏导数是： \[ \frac{\partial z}{\partial x} = \frac{1}{xy} \cdot y = \frac{1}{x} \] 函数 \( z \) 关于 \( y \) 的一阶偏导数是： \[ \frac{\partial z}{\partial y} = \frac{1}{xy} \cdot x = \frac{1}{y} \] 接下来，计算它们的二阶偏导数和三阶偏导数。 对 \( \frac{\partial z}{\partial x} \) 关于 \( x \) 求二阶偏导数： \[ \frac{\partial^2 z}{\partial x^2} = -\frac{1}{x^2} \] 对 \( \frac{\partial^2 z}{\partial x^2} \) 关于 \( y \) 求三阶偏导数： \[ \frac{\partial^3 z}{\partial x^2 \partial y} = 0 \] 对 \( \frac{\partial z}{\partial y} \) 关于 \( y \) 求二阶偏导数： \[ \frac{\partial^2 z}{\partial y^2} = -\frac{1}{y^2} \] 对 \( \frac{\partial^2 z}{\partial y^2} \) 关于 \( x \) 求三阶偏导数： \[ \frac{\partial^3 z}{\partial x \partial y^2} = 0 \] 由于 \( \frac{\partial^3 z}{\partial x^2 \partial y} \) 和 \( \frac{\partial^3 z}{\partial x \partial y^2} \) 都等于 0，所以： \[ \frac{\partial^3 z}{\partial x^2 \partial y} - \frac{\partial^3 z}{\partial x \partial y^2} = 0 - 0 = 0 \] 因此，所求的差值是 0。
Second Derivative of a Complicated Function
Câu hỏi yêu cầu tìm đạo hàm cấp hai của hàm số \( y = \frac{x + 3}{2x - 1} - \frac{1}{\sin x} \). Đầu tiên, chúng ta cần tìm đạo hàm cấp một của hàm số này. Đạo hàm của hàm số \( y = \frac{x + 3}{2x - 1} \) sử dụng quy tắc của hàm phân thức (quy tắc chia) là: \( y' = \frac{(2x - 1)\cdot1 - (x + 3)\cdot2}{(2x - 1)^2} = \frac{-x - 5}{(2x - 1)^2} \) Đạo hàm của hàm số \( y = -\frac{1}{\sin x} \) là: \( y' = -\frac{\cos x}{(\sin x)^2} \) sử dụng quy tắc đạo hàm của hàm lượng giác kết hợp với quy tắc chung về đạo hàm của hàm phân thức. Tiếp theo, để tìm đạo hàm cấp hai, ta tiếp tục lấy đạo hàm của \( y' \): Đối với phần \( \frac{-x - 5}{(2x - 1)^2} \), đạo hàm cấp hai sẽ là: \( y'' = \frac{\begin{bmatrix}(2x - 1)^2\cdot(-1) - (-x - 5)\cdot2\cdot2\cdot(2x - 1)\end{bmatrix}}{(2x - 1)^4} \) \( y'' = \frac{- (2x - 1)^2 + 4(x + 5)(2x - 1)}{(2x - 1)^4} \) \( y'' = \frac{- (2x - 1)^2 + 8x^2 + 36x - 20}{(2x - 1)^4} \) \( y'' = \frac{4x^2 + 36x - 18}{(2x - 1)^4} \) Đối với phần \( -\frac{\cos x}{(\sin x)^2} \), đạo hàm cấp hai sẽ là: \( y'' = -\frac{\begin{bmatrix}(-\sin x)(\sin x)^2 - (\cos x)\cdot2\cdot(\sin x)\cdot\cos x\end{bmatrix}}{(\sin x)^4} \) \( y'' = -\frac{-(\sin x)^3 - 2(\cos x)^2\cdot(\sin x)}{(\sin x)^4} \) \( y'' = -\frac{-(\sin x)^3 - 2(1 - (\sin x)^2)(\sin x)}{(\sin x)^4} \) \( y'' = -\frac{-(\sin x)^3 - 2\sin x + 2(\sin x)^3}{(\sin x)^4} \) \( y'' = -\frac{(sin x)^3 + 2(\sin x)^3 -2\sin x}{(\sin x)^4} \) \( y'' = -\frac{3(\sin x)^3 -2\sin x}{(\sin x)^4} \) \( y'' = -\frac{3\sin x - 2}{(\sin x)^3} \) Cuối cùng, ta cộng đạo hàm cấp hai của hai phần lại với nhau: \( y'' = \frac{4x^2 + 36x - 18}{(2x - 1)^4} - \frac{3\sin x - 2}{(\sin x)^3} \) Đó là đạo hàm cấp hai của hàm số ban đầu.
Finding Differential Equation for General Solution
The image contains a handwritten request to find the differential equation associated with the general solution `y = Be^{(αx + β)}`, where α, β, and B are constants. To find the differential equation, we need to eliminate the constants α, β, and B from the given general solution. Given: \[ y = Be^{(αx + β)} \] Step 1: Differentiate both sides with respect to x. \[ \frac{dy}{dx} = Be^{(αx + β)} \cdot α \] Step 2: We need a second derivative since there are two independent constants (α and β), so differentiate once more with respect to x. \[ \frac{d^2y}{dx^2} = \frac{d}{dx} \left( Be^{(αx + β)} \cdot α \right) = Be^{(αx + β)} \cdot α^2 \] Step 3: Now, we need to eliminate B from these equations. From the first derivative, isolate B: \[ B = \frac{1}{α} \frac{dy}{dx} \cdot e^{-(αx + β)} \] Step 4: Substituting B from the equation obtained from the first derivative into the second derivative, we get: \[ \frac{d^2y}{dx^2} = \frac{1}{α} \frac{dy}{dx} \cdot e^{-(αx + β)} \cdot α^2 \cdot e^{(αx + β)} \] Step 5: Cancel out the exponential terms \( e^{-(αx + β)} \cdot e^{(αx + β)} \) which is equal to 1. This gives us: \[ \frac{d^2y}{dx^2} = α \frac{dy}{dx} \] Step 6: We still need to eliminate α and β. Since β does not directly appear in the derived differential equation, we're left with only eliminating α. To eliminate α, we observe that our first derivative is: \[ \frac{dy}{dx} = α y \] Step 7: Divide the second derivative by y: \[ \frac{1}{y} \frac{d^2y}{dx^2} = α^2 \] Step 8: From the relation in Step 6 where we have \( \frac{dy}{dx} = α y \), we can write α as: \[ α = \frac{1}{y} \frac{dy}{dx} \] Step 9: Substituting this value of α back into the Step 7 equation: \[ \frac{1}{y} \frac{d^2y}{dx^2} = \left( \frac{1}{y} \frac{dy}{dx} \right)^2 \] Step 10: This is the desired second-order homogeneous differential equation with constant coefficients, and it is free from the constants α, β, and B: \[ \frac{d^2y}{dx^2} = \left( \frac{dy}{dx} \right)^2 / y \] Or equivalently, \[ y \frac{d^2y}{dx^2} - \left( \frac{dy}{dx} \right)^2 = 0 \] This is the differential equation associated with the given general solution.
Finding Second Derivative of Natural Logarithm Function
The image shows a mathematical expression and a request to find the second derivative of the function with respect to x. The function is \( y = \ln(2x^2 + x) \). We need to find \( \frac{d^2y}{dx^2} \) when \( x = -1 \). First, let's find the first derivative \( \frac{dy}{dx} \). To do this, we'll use the chain rule since y is the natural logarithm of a function of x. Let \( u = 2x^2 + x \), so \( y = \ln(u) \). First, let's differentiate \( u \) with respect to \( x \): \[ \frac{du}{dx} = \frac{d(2x^2 + x)}{dx} = 4x + 1 \] Now let's differentiate \( y \) with respect to \( u \): \[ \frac{dy}{du} = \frac{d(\ln(u))}{du} = \frac{1}{u} \] Now use the chain rule to find \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{u} \cdot (4x + 1) = \frac{4x + 1}{2x^2 + x} \] Next, we'll find the second derivative \( \frac{d^2y}{dx^2} \). For this, we need to differentiate \( \frac{dy}{dx} \) with respect to \( x \) using the quotient rule: Let \( v = 4x + 1 \) and \( w = 2x^2 + x \), so \( v' = 4 \) and \( w' = 4x + 1 \). Now apply the quotient rule \( \left( \frac{v}{w} \right)' = \frac{v'w - vw'}{w^2} \): \[ \frac{d^2y}{dx^2} = \frac{v'w - vw'}{w^2} = \frac{4(2x^2 + x) - (4x + 1)(4x + 1)}{(2x^2 + x)^2} \] Now simplify and find \( \frac{d^2y}{dx^2} \) when \( x = -1 \): \[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - (16x^2 + 8x + 1)}{(2x^2 + x)^2} \] \[ \frac{d^2y}{dx^2} = \frac{8x^2 + 4x - 16x^2 - 8x - 1}{(2x^2 + x)^2} \] \[ \frac{d^2y}{dx^2} = \frac{-8x^2 - 4x - 1}{(2x^2 + x)^2} \] Now plug in \( x = -1 \): \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8(-1)^2 - 4(-1) - 1}{(2(-1)^2 - 1)^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-8 + 4 - 1}{(2 + (-1))^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = \frac{-5}{1^2} \] \[ \frac{d^2y}{dx^2}\Bigg|_{x=-1} = -5 \] The second derivative of the function at \( x = -1 \) is \( -5 \).
CamTutor
In regards to math, we are professionals.
Get In Touch
Email: camtutor.ai@gmail.com