(Hint: Work with the right side. Add and subtract f(x)g(x + h) in the numerator.) 相关知识点: 试题来源: 解析Sample answer:[f(x)-g(x)]'=limlimits_(h→0)(f(x+h)g(x+h)-f(x)g(x))h=limlimits_(h→0)(f(x+h)g(x+h)-f(x)g(x+h)+f(x)g(x+h)-f(x)g(x))h...
结果1 题目 Use the Product Rule twice to prove that if f, g, and h are differentiable functions of x, then(dx)[f(x)g(x)h(x)]=f'(x)g(x)h(x)+f(x)g'(x)h(x)+f(x)g(x)h'(x). 相关知识点: 试题来源: 解析 f(x)=3x^2-2b-1 反馈 收藏 ...
Chapter 10/ Lesson 14 214K Understand what the product rule is. Learn about the product rule in calculus. Know about the derivative multiplication rule and the product rule equation. Explore our homework questions and answers library Search ...
To use the product rule do we have to write the cross product as followed?? $\overrightarrow{\sigma}(t)=(x_1(t), x_2(t), x_3)), \overrightarrow{\rho}(t)=(y_1(t), y_2(t), y_3(t))$ $$\overrightarrow{\sigma}(t) \times \overrightarrow{\rho}(t)=\begin{vmat...
Suppose that f(x,y) and g(x,y) are two functions of x and y then product rule for the derivative is defined by {eq}\displaystyle \frac{\partial }{{\partial x}}(f(x,y) \cdot g(x,y)) = f(x,y)\frac{{\partial g(x,y)}}{{\partial x}} + g(x,y)\frac{{\partial f(...
Use the limit definition of the derivative to prove the followingspecial case of the Product Rule:d/(dx)(xf(x))=xf'(x)+f(x) 相关知识点: 试题来源: 解析 d/(dx)(xf(x))=lim_(h→0)((x+h)f(x+h)-f(x))/ =lim_(h→0)(x(f(x+h)-f(x))/h+f(x+h)) =xlim_(h→0)(...
In General, the factory will provide to customer a positive, to prove their products 翻译结果4复制译文编辑译文朗读译文返回顶部 As a general rule, the plant will be provided to the customer, the active product to demonstrate that they 翻译结果5复制译文编辑译文朗读译文返回顶部 Generally speaking, ...
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