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## Calculus, all content (2017 edition)

### Course: Calculus, all content (2017 edition)>Unit 2

Lesson 19: Product rule

# Worked example: Product rule with mixed implicit & explicit

We explore how to calculate the derivative of F(x) = f(x)⋅g(x) at x = -1, given the values of f and f' at x = -1 and g(x) = 1/x. By applying the product rule, we efficiently determine F'(x) and evaluate it at the specified point.

## Want to join the conversation?

• Since we have g(-1) = -1, so why doesn't taking derivative of a constant i.e g'(-1) = 0? Where am I wrong?
(1 vote)
• Just because g(-1) = -1 does not make g a constant function! All g(-1) = -1 tells us is that g has value -1 specifically when x is -1.
• As for the title of this video, "Product rule with mixed implicit & explicit", is the function f implicit?
• In math, an explicit function is simply a function where the dependent variable is given explicitly; you don't have to algebraically manipulate the function to know what the dependent variable is. Every y=f(x) is an explicit function because it is clear that the value of y is dependent on the value of x.
On the other side, an implicit function is any "function" where there doesn't appear to be any dependent variable, such as x^2+y^2=1. Later on, you will likely learn about implicit differentiation, in which you calculate the slope of a curve given by any an implicit function, rather than just taking the derivative of an explicit function like we are doing currently.

It appears that neither f nor g is defined implicitly; they're both explicit, so the title of this video isn't helpful.