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## Differential Calculus

### Course: Differential Calculus>Unit 2

Lesson 10: Product rule

# Worked example: Product rule with mixed implicit & explicit

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.B (LO)
,
FUN‑3.B.1 (EK)
Given the values of f and f' at x=-1 and that g(x)=1/x, Sal evaluates the derivative of F(x)=f(x)⋅g(x) at x=-1.

## 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.