Zack K said:My issue is that I do not know what h(0) is so I can't plug that into my equation to find h'0
Then h(x) is equal to 2/π?verty said:You are given that ##h(x) + x\ \cos(h(x)) = x^2 + 3x + {2 \over \pi}##. This is true whatever we make x. What happens when x = 0?
No, ##h(0) = \frac 2 \pi##. You don't have an explicit formula for h(x).Zack K said:Then h(x) is equal to 2/π?
Zack k said:I differentiated both sides, giving h'(x)+cos(h(x))-xh'(x)sin(h(x))=2x+3
The value of an implicit derivative is a numerical value that represents the rate of change of a function with respect to its independent variable. It is calculated using the derivative rules and can be used to determine the slope of a tangent line at a specific point on a curve.
The value of an implicit derivative is calculated by taking the derivative of both sides of an implicit equation with respect to the independent variable. This involves using the chain rule and product rule, if necessary, to simplify the equation and solve for the implicit derivative.
The value of an implicit derivative is significant because it provides information about the behavior of a function at a specific point. It can be used to determine if a function is increasing or decreasing at that point, as well as the slope of the tangent line. This information is useful in many real-world applications, such as in physics, economics, and engineering.
An implicit derivative is calculated using an implicit equation, where the dependent variable is not explicitly written in terms of the independent variable. In contrast, an explicit derivative is calculated using an explicit equation, where the dependent variable is written explicitly in terms of the independent variable. The value of an implicit derivative is not always easy to find, while the value of an explicit derivative can be easily calculated using the power rule.
Yes, the value of an implicit derivative can be negative. This means that the function is decreasing at that point. However, the value of an implicit derivative can also be positive or zero, depending on the behavior of the function at that point. It is important to consider the sign of the implicit derivative when interpreting its value in real-world situations.