# Chain rule? Calculus ?

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

Relevance

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)

:)>

• 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).

• (∂y/∂x)=(∂y/∂u)(∂u/∂x)

(∂u/∂x)=4e^(4x)

(∂y/∂u)= -60 sin(6 u)

• 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

• 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) )

• 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))