The chain rule tells us how to find the derivative of a composite function. Brush up on your knowledge of composite functions, and learn how to apply the chain rule correctly.
Log in Anto Q 7 years agoPosted 7 years ago. Direct link to Anto Q's post “Why are sin(x), cos(x), e...” Why are sin(x), cos(x), etc, called transcendental functions? • (37 votes) andrewp18 7 years agoPosted 7 years ago. Direct link to andrewp18's post “They cannot be expressed ...” They cannot be expressed as finite-degree polynomials. (70 votes) jade jericho 6 years agoPosted 6 years ago. Direct link to jade jericho's post “On Problem 2. I chose the...” On Problem 2. I chose the answer "not a composite function" because i took cos^2(x) to be cos(x)*cos(x) and hence used the product rule to differentiate. Is that okay? • (34 votes) checedrodgers 3 years agoPosted 3 years ago. Direct link to checedrodgers's post “Yes, applying the chain r...” Yes, applying the chain rule and applying the product rule are both valid ways to take a derivative in Problem 2. The placement of the problem on the page is a little misleading. Immediately before the problem, we read, "students often confuse compositions ... with products". This suggests that the problem we are about to work (Problem 2) will teach us the difference between compositions and products, but, surprisingly, cos^2(x) is both a composition _and_ a product. You can see this by plugging the following two lines into Wolfram Alpha (one at a time) and clicking "step-by-step-solution": For d/dx sin(x)cos(x), W.A. applies the product rule. For d/dx cos(x)cos(x), W.A. recognizes that we can rewrite as a composition d/dx cos^2(x) and apply the chain rule. In summary, there are some functions that can be written only as compositions, like d/dx ln(cos(x)). There are other functions that can be written only as products, like d/dx sin(x)cos(x). And there are other functions that can be written both as products and as compositions, like d/dx cos(x)cos(x). (15 votes) Hima Praveen 7 years agoPosted 7 years ago. Direct link to Hima Praveen's post “how do you know when to u...” how do you know when to use chain rule? • (10 votes) Howard Bradley 7 years agoPosted 7 years ago. Direct link to Howard Bradley's post “One uses the chain rule w...” One uses the chain rule when differentiating a function that can be expressed as a function of a function. (21 votes) 20leunge 7 years agoPosted 7 years ago. Direct link to 20leunge's post “Why is it called the chai...” Why is it called the chain rule? • (11 votes) Moon Bears 7 years agoPosted 7 years ago. Direct link to Moon Bears's post “It is called the chain ru...” It is called the chain rule, because of the formula for the chain rule when considering differentiable functions from R^n to R^m. Read more at wikipedia https://en.wikipedia.org/wiki/Chain_rule#Higher_dimensions (11 votes) Ryan 6 years agoPosted 6 years ago. Direct link to Ryan's post “Why is the derivative of ...” Why is the derivative of 3x = 3, but the derivative of 3(sin x) = 3(cos x)? • (4 votes) 6 years agoPosted 6 years ago. Direct link to cossine's post “The derivative gives you ...” The derivative gives you a rate of change or to put it more simply the gradient of a function. Since the rate change varies it is not the same. Like 3x =y, has a constant gradient of 3. But 3sinx has a gradient of 3cos(x) which depends on x which is not constant. (12 votes) colinhill a year agoPosted a year ago. Direct link to colinhill's post “One thing that has been a...” One thing that has been a little bit shaky with me is what can we define as its own function. I understand cos(x^2) being defined as f(x) = cos(x) and g(x) = x^2, but can we say cos(x) is also a composite function? In that example,f(x) = cos(x) and g(x) = x then we solve with the chain rule. If we can, I'm assuming it's redundant, but I'm just looking for a "line" of when we can and can't define new functions. Also, can we combine different "instances" (sorry for the poor terminology) of variables as ONE unique function? For example, if we had f(x) = (x^2)(cos(x))(sin(x)), could we combine (x^2) and cos(x) to be one function? so, g(x) = (x^2)(cos(x)) and h(x) = sin(x). If so, then could we apply the product rule with those two functions (one of which being a combination of both (x^2) and cos(x))? So, f(x) = (g(x))(h(x)) Sorry about the long question, I am just hoping for a more clear picture of when we can assign variables as their own function. Thanks. • (6 votes) kubleeka a year agoPosted a year ago. Direct link to kubleeka's post “That's the beauty of it: ...” That's the beauty of it: we can describe variables as separate functions however we like. We can say that x² is the function f(x)=x² composed with g(x)=x, and applying chain rule gives us Or we can rewrite x as e^(ln(x)). Then chain rule gives the derivative of x as For your product rule example, yes we could consider x²cos(x) to be a single function, and in fact it would be convenient to do so, since we only know how to apply the product rule to products of two functions. By doing this, we find the derivative to be d/dx[x²cos(x)]·sin(x)+(x²cos(x))·cos(x) and now we can simplify this by computing the derivative of x²cos(x) using the product rule again. There is no "line". We can divvy up expressions, introduce multiplications by 1, or write simple variables as compositions of inverse functions however we like, however makes the problem more convenient to solve. (8 votes) Liang a year agoPosted a year ago. Direct link to Liang's post “can u not see [cos (x)]^2...” can u not see [cos (x)]^2 as a composite function, but see it as: cos (x)*cos (x), and use the product rule to find the derivative (using both chain rule and product rule ends up the same derivative)? thx! • (6 votes) Venkata a year agoPosted a year ago. Direct link to Venkata's post “Yep! You can do that. And...” Yep! You can do that. And as you said, you'll get the same answer in both cases (-2sin(x)cos(x)) (7 votes) vishnuprashanth.cvi 7 years agoPosted 7 years ago. Direct link to vishnuprashanth.cvi's post “Can someone kindly help m...” Can someone kindly help me on how to differentiate (x-11)^3 / (x+3) using the chain rule? • (1 vote) kubleeka 7 years agoPosted 7 years ago. Direct link to kubleeka's post “You'll need quotient or p...” You'll need quotient or product rule in addition to the chain rule. First let's find the derivative of (x-11)³. Outer function is x³, inner function is x-11. So we take Now, the derivative of x+3 is just 1. Putting these together with the quotient rule, we get (12 votes) hmc 5 years agoPosted 5 years ago. Direct link to hmc's post “This is great. Is there a...” This is great. Is there an article showing the chain rule when there are more than 2 functions in a composite? • (0 votes) The #1 Pokemon Proponent 5 years agoPosted 5 years ago. Direct link to The #1 Pokemon Proponent's post “You can perfectly extend ...” You can perfectly extend what applies for 2 functions to 3 functions or 10 functions or 100000 functions. For 2 functions (f(g(x)))'= f'(g(x))*g'(x) for 3 functions (f(g(h(x))))' = f'(g(h(x)))*g'(h(x)) *h'(x) and so on. (10 votes) Revant Sai T 5 years agoPosted 5 years ago. Direct link to Revant Sai T's post “Can you please make a vid...” Can you please make a video on how to find the minimum value of 2^sin(x) + 2^cos(x). Also, what is the inner function and outer function for 2^sin(x)?? • (2 votes) Alex 5 years agoPosted 5 years ago. Direct link to Alex's post “While the AM-GM inequalit...” While the AM-GM inequality is useful, I'm just going to do it as a standard optimization problem. Because you asked, inner function is sin(x) and outer function is 2^x. f(x) = 2^(sin x) + 2^(cos x) min(f(x)) = f(5pi / 4 + 2pi * k) = 1.225. (5 votes)Want to join the conversation?
https://en.wikipedia.org/wiki/Transcendental_function
d/dx sin(x)cos(x)
d/dx cos(x)cos(x)
eg sin(x²) or ln(arctan(x))
f'(g(x))·g'(x)
g'(x)=1, and g(x)=x, so this simplifies to
f'(x)
which is 2x.
e^(ln(x))·(1/x), or x/x, or 1.
TIA
d/dx((x-11)³)
d/(x-11) (x-11)³ •d/dx (x-11)
3(x-11)²•1
3(x-11)²
d/dx ((x-11)³/(x+3))
[(x+3)•d/dx(x-11)³ -(x-11)³•d/dx(x+3)]/((x+3)²)
Substitute the derivatives that we know and we get
[(x+3)•3(x-11)² -(x-11)³•1]/((x+3)²)
This is our answer. To simplify, we can factor an (x-11)² out of the numerator and get
(x-11)²•(3x+9 -(x-11))/((x+3)²)
(x-11)²•(2x+20)/((x+3)²)
Thanks, any help would be appreciated
f'(x) = ln(2) * 2^(sin x) * cos x - ln(2) * 2^(cos x) * sin x
This is undefined nowhere, so we look for points where f'(x) = 0, or 2^(sin x) cos x = 2^(cos x) sin x; this expression is (essentially) equivalent to sin(x) = cos(x), which is true at pi/4 + k * pi.
The minimum value of f should be at 5pi/4, where (sin x, cos x) = (-sqrt(2) / 2, -sqrt(2) / 2).
