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

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

Key Concepts

Derivative of Trigonometric Functions

Differentiating trigonometric functions is a key aspect of calculus, involving specific rules for functions like sine, cosine, and their reciprocals. Recognizing the derivatives of these functions, such as the derivative of the secant squared function or cosine function, is important in problems that require the application of the chain rule to trigonometric expressions.

Evaluation of the Derivative at a Specific Point

After finding the general derivative of a function, it is often necessary to evaluate it at a particular point. This concept involves substituting the specified value into the derivative expression to find the rate of change of the function at that exact point, which is a common practice in calculus to understand the behavior of functions at given inputs.

Chain Rule

The chain rule is a fundamental differentiation technique used when dealing with composite functions. It states that if you have a function composed of one function inside another, such as f(g(x)), then its derivative can be found by multiplying the derivative of the outer function evaluated at the inner function by the derivative of the inner function.

Composite Functions

Composite functions occur when one function is applied to the result of another function. Understanding how to decompose and recompose these functions is essential for applying differentiation techniques like the chain rule. This concept is widely used in calculus especially when functions are nested within one another.

Transcript

00:01 Section 3 .684, we're dealing with composition functions and chain rule.

00:07 So we're asked here to find f composition g prime at x equal one -fourth.

00:14 So what we're relying on here is the derivative of the composition is f prime of g of x times g prime of x, and that is our chain rule.

00:25 So this function in f is we have what f of u is equal to u.

00:33 Plus the cosine of u to the minus two power.

00:38 So what is the derivative of f? that is going to be 1 plus minus 2 cosine to the minus 3 of u.

00:49 And then what's the derivative of cosine is minus sine of u? so this is going to be 1 plus 2 sine of u over cosine cubed of u.

01:06 So let me make that a little bit clearer so 1 plus 2 sign of u so 1 plus 2 sine of u over the cosine cubed of you so that is f prime at u now if we were to substitute the value of g which is pi x into that that is 1 plus 2 sine of pi x over cosine cubed cubed of pi x and then the derivative of g with respect to x is simply pi so the composition is formed by multiplying these two expressions together so i know that f of g of x prime is going to be equal to pi times 1 plus 2 sine of pi x over the cosine cubed of pi x and originally i was tasked at evaluating this when x is equal to one -fourth so let's evaluate this when x is equal to one -fourth so that is going to give us pi times one plus two the sign of pi when you substitute x equal to one -fourth that's the sign of pi over four so that is the square root of 2 over 2 divided by the cosine cubed and so let's just take that out the cosine so the same thing cosine of pi over 4 is the same thing square of 2 over 2 so this is the square root of 2 over 2 cubed so what do i have here i've got this is pi and then it looks like i've got 1 plus this is the square root of 2 and now let's look can see what is this square root of two over two cubed let's figure out what that is so the square root of two over two cubed that is going to be two to the three halves over two to the third which is two to the three halves over two to the six halves which is one over two to the three halves which is two to the minus three halves so this this is 2 to the minus 3 halves...

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