Chain rule? Calculus ?

If y=f(u)=10cos(6u) and u=g(x)=e^4x, then by the chain rule,

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  • 1 month ago
    Favorite Answer

    Using u = e⁴ˣ, we have y = 10cos(6u) 

    so, du/dx = 4e⁴ˣ and dy/du = -60sin(6u)

    Now, dy/dx = (dy/du).(du/dx) so,

    dy/dx = (4e⁴ˣ)(-60sin(6u))

    so, dy/dx = -240e⁴ˣsin(6u)

    Now, as u = e⁴ˣ we have:

    dy/dx = -240e⁴ˣsin(6e⁴ˣ)

    :)>

  • 1 month ago

    Do you want to find dy/dx? If so, then

    dy/dx=

    [dy/du][du/dx]=

    [-60sin(6u)][4e^(4x)]=

    -960sin[6e^(4x)]e^(4x).

  • 1 month ago

    (∂y/∂x)=(∂y/∂u)(∂u/∂x)

    (∂u/∂x)=4e^(4x)

    (∂y/∂u)= -60 sin(6 u)

  • ?
    Lv 7
    1 month ago

    Given y = f(u) = 10 cos 6u    and   u = g(x) = e ⁴ˣ

    y =  10 cos 6 e ⁴ˣ

    y ' = 10 • 6 sin 6e ⁴ˣ • 4 e ⁴ˣ

    y ' = 240 e ⁴ˣ  sin 6e ⁴ˣ.............ANS

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  • 1 month ago

    i am assuming you meant g(x) = e^(4x) instead of g(x) = xe^4 

    y = 10cos(6u)

    u = e^(4x) 

    dy/du = -60 sin(6u)

    du/dx = 4e^(4x)

    dy/dx = dy/du * du/dx

    dy/dx = [ -60 sin( 6u ) ] [ 4e^(4x) ]

    dy/dx = -240 e^(4x) sin( 6 *  e^(4x) ) 

    dy/dx = -240 e^(4x) sin( 6e^(4x) ) 

  • 1 month ago

    Assuming you want the derivative:

    y = 10cos(6e⁴ˣ)

    d/dx(10u) = 10(du/dx)

    d/dx(cos(u)) = -sin(u)(du/dx)

    d/dx(e⁴ˣ) = 4e⁴ˣ

    y' = -10sin(6e⁴ˣ)(6)(4e⁴ˣ)

    y' = -240(e⁴ˣ)(sin(6e⁴ˣ))

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