Find the value of (f ∘g)^' at the given value of x. f(u)=(2 u)/(u^2+1), u=g(x)=10 x^2+x+1, x=0

Name: Problem 85, Chapter 3: University Calculus: Early Transcendentals
Uploaded: 2020-05-05T17:30:16-08:00
Duration: 5 min
Channel: Robert Daugherty
Description: Find the value of (f ∘g)^' at the given value of x. f(u)=(2 u)/(u^2+1), u=g(x)=10 x^2+x+1, x=0

step 1 Section 3 .685. So we're dealing with composition functions. They're asking us to find the deriv

Find the value of (f ∘g)^' at the given value of x. f(u)=(2...

Key Concepts

Differentiation of Rational Functions

Differentiating rational functions typically involves applying the quotient rule, which is used when the function is expressed as a ratio of two differentiable functions. In many problems, such as when the outer function in a composite function is a rational function, understanding how to correctly compute its derivative is critical to applying the chain rule effectively.

Chain Rule

The chain rule is a core principle in calculus used to differentiate composite functions. It states that the derivative of a composite function f(g(x)) is the product of the derivative of the outer function evaluated at the inner function, f'(g(x)), and the derivative of the inner function, g'(x). This rule allows one to systematically break down and compute the derivative of complex function compositions.

Composite Functions

Composite functions are formed when one function is applied to the result of another function, typically noted as (f ? g)(x) = f(g(x)). This concept is essential because it combines the effects of two or more functions into one operation, and understanding it is key when dealing with problems that require evaluating or differentiating such compositions.

Transcript

00:01 Section 3 .685.

00:04 So we're dealing with composition functions.

00:06 They're asking us to find the derivative of f composition g at the point x equals 0.

00:15 Now we know that the chain rule is what helps us to find the derivative of composition.

00:20 So it's going to be f prime, evaluated at g of x times g prime of x.

00:26 So let's just go find those derivatives.

00:29 So what is the derivative of f with respect to you? well that's going to be our handy quotient rule so that is going to be the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator over the denominator squared and so when we simplify this we're going to get it looks like let's just go ahead and work on this you're going to have in this numerator here you're going to have what 2 u squared plus 2 so 2 u squared plus 2 minus 4 u squared so that's just going to be 2 minus 2 u squared so it's going to be 2 times um so if you factor out of 2 there it's going to be 2 1 minus u squared so let's just go ahead and work on that so that's 2 1 minus u squared so i end up with 2 1 minus u squared over u squared plus 1 squared.

01:53 And so now if i substitute g into that equation, that's going to give me 2 times 1 minus, and g is what 10 x squared plus x plus 1 squared? so that's 2 times 1 minus u squared over, and then you're going to have 10 x squared plus x plus 1 plus 1 squared.

02:23 Plus 1 squared and then the derivative of g that's the easy one that's going to be 20x plus 1 so my composition is going to be the product of these two so let's just go to another screen and we'll write all of that down so f composition g is going to be equal to 2 times 1 minus so it's going to be 2 times 1 minus that whole quantity 10x squared plus x plus 1 squared over 10x squared plus x plus 1 squared plus 1 times that derivative which is what 20x plus 1 okay and what we're asked to do now we were asked to originally evaluate this thing at x equal to 0 so i've got to evaluate this at x equal to zero so when x is equal to zero, this turns into two, one minus, and then you're going to have what, one squared over, and then you're going to have when you put zero on the bottom, you're going to have one squared plus one squared.

04:10 So what i see immediately is this one minus one...

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