Calc help, chain rule?

Question

y = x [sec (3 - 8x)]
Find dy/dx.
I would greatly appreciate if someone could help me figure this out, and point me to a good site that has an in-depth explanation of all of the possible cases involving the chain rule. After all, getting the answer to this question without knowing how it was arrived at will not help me understand the subject material...

Answer 1

y = x / [ cos (3 - 8x) ]
Using quotient rule:-
Numerator, N = cos(3 - 8x) - x sin(3 - 8x) (- 8)
N = cos (3 - 8x) + 8x sin (3 - 8x)

Denominator, D = cos ² (3 - 8x)

N / D
= 1 / cos (3 - 8x) + 8x tan (3 - 8x) sec(3 - 8x)
= sec(3 - 8x) [ 1 + 8x tan (3 - 8x) ]

Answer 2

d(x sec (3-8x))/dx

Before you even think about the chain rule, remember that you have the product of two functions, thus the first step is to apply the product rule:

dx/dx sec (3-8x) + x d(sec (3-8x))/dx

Of course, dx/dx=1, so we have:

sec (3-8x) + x d(sec (3-8x))/dx

NOW, we apply the chain rule. Since d(sec x)/dx = sec x tan x, we have:

sec (3-8x) + x sec (3-8x) tan (3-8x) d(3-8x)/dx

The last polynomial differentiates easily:

sec (3-8x) + x sec (3-8x) tan (3-8x) (-8)

So simplifying the result:

sec (3-8x) - 8x sec (3-8x) tan (3-8x)
sec (3-8x) (1 - 8x tan (3-8x))

And we are done.

Really, there's only one "case" of the chain rule, and that is when you have one function "inside" another function (the technical term is composed with). The key is to recognize that you need to work from the outside in -- that is, don't worry about applying the chain rule until you have just the composition of functions to work with. If you see something else outside of that, such as a multiplication, you need to use the appropriate rule on that first, then apply the chain rule.

Answer 3

Technically, the chain rule ALWAYS takes place whenever you are taking the derivative of any function.

For example:
Find dy/dx of 3x^2
dy/dx = 6x * d/dx and since the derivative of x = 1 then we simply ignore it because it wouldn't change the answer.

The only times when the chain rule actually changes an answer is when the derivative of the inside function doesn't equal 1. In other words you have something else besides simply x inside.

Example:
Find dy/dx of cos(2x) take the derivative of the outside function first then inside function.
dy/dx = -sin(2x) * d/dx in this case d/dx = 2
dy/dx = -2sin(2x)

Now for your problem.

y = x [sec (3 - 8x)] you first need to apply the product rule and then use the chain rule.

dy/dx = u'v + uv'
u = x
v=sec (3 - 8x)

dy/dx = 1*sec (3 - 8x) + x[sec (3 - 8x)*tan(3 - 8x)] * -8 simplify
dy/dx = sec (3 - 8x) + -8x*sec (3 - 8x)*tan(3 - 8x) [ans]

Answer 4

howdy thats no longer the appropriate thank you to do it, and that i understand your confusion.. the order in which you do those issues would properly be somewhat confusing.. even nevertheless it helps to get your straightforward strategies at modern only you're able to need to prepare the chain rule in basic terms once you're DIFFERENTIATING some thing as an occasion, on your question, prepare the product rule first product rule calls which you would be able to maintain one functionality as this is, jointly as differentiating the different, and then taking their product.. after which you shop the different functionality as this is, jointly as differentiating the different, and then multiply the two after which you upload the two words you acquire now the place does the chain rule are available?? u use the chain rule interior the section the place you DIFFERENTIATE between the applications on your question w(x) = (x - a million)^12 (x + 2)^10 , you first shop (x - a million)^12 as this is, and differentiate (x + 2)^10 , and then multiply, to get [ (x-a million)^12 ] * [ 10(x+2)^9 * d/dx(x-a million) ] (did u word the place the chain rule became into utilized?) and then next, you shop (x + 2)^10 as this is, and differentiate (x - a million)^12 to get [ (x + 2)^10 ] * [ 12(x-a million)^11 * d/dx(x-a million) ] after which you upload the two words to get w'(x) = [ (x-a million)^12 ] * [ 10(x+2)^9 * d/dx(x-a million) ] + [ (x + 2)^10 ] * [ 12(x-a million)^11 * d/dx(x-a million) ] w'(x) = [ (x-a million)^12 ] * [ 10(x+2)^9 ] + [ (x + 2)^10 ] * [ 12(x-a million)^11 ]

Answer 5

y = x [ sec ( 3 - 8x ) ]

Here, y is a product of two functions, x and sex ( 3 - 8x )

Differentiation of product of two functions f(x) and g(x) is
f(x)*g'(x) + g(x)*f'(x)

Hence, dy/dx = x * d/dx sec ( 3 - 8x ) + sec ( 3 - 8x ) * d/dx (x)
= x d/dx sec ( 3 - 8x ) + sec ( 3 - 8x ) ... ... ( 1 )

Now, for differentiation of sec ( 3 - 8x ), we use the chain rule.
For this, we first differentiate sec ( 3 - 8x ) using the formula for differntiation of sec x and then multiply the result with differentiation of 3 - 8x.

Thus, d/dx sec ( 3 - 8x ) = sec ( 3 - 8x ) tan ( 3 - 8x ) d/dx ( 3 - 8x)
= - 8 sec ( 3 - 8x ) tan ( 3 - 8x )

Putting in ( 1 ),

dy/dx = - 8x sec ( 3 - 8x ) tan ( 3 - 8x ) + sec ( 3 - 8x )

Answer 6

Hi,

First off, you'll actually need a variant of the chain rule called thr product rule for this one.

f(x) = u*v
df/dx = (du/dx)v + u(dv/dx)

so you'll have
u = x
v = sec(3-8x)
du/dx = 1
dv/dx = d/dx(1/sin x ) = -8{cos(3-8x)/sin(3-8x)^2} (See the attached page)

so
df/dx = (du/dx)v + u(dv/dx)
df/dx = (1)(sec(3-8x)) + x (-8{cos(3-8x)/sin(3-8x)^2})

Hope that helps,
Matt

Answer 7

Let...

g(x) = x
h(x) = sec(x)
i(x) = 3 - 8x

then y = g(x) * (h o i) (x) h o i, h composed of i

in order to obtain dy/dx use the product rule and then the chain rule. and thats it