How do you use the chain rule to differentiate #y=cos(sqrt(8t+11))#?

Question

Charles Arocha · Accepted Answer

Answer is #-4sin(sqrt(8t+11))/((sqrt(8t+11)))#

Differentiating y with respect to t.
#dy/dt=d/dt (cos(sqrt8t+11)) #

Using Chain Rule

#(f(g))' = f'(g)* g'#

Let g = #sqrt(8t+11)#
New expression will be #d/(dg) (cos(g))# i.e. Differentiated with respect to g.

#d/(dg)(cos(g))*d/(dt)(sqrt(8t+11))#.....................(1)
Using #d/(dx)(cos (x))=-sin(x)# solve the equation (1)

#= -sin(g)*d/(dt)(sqrt(8t+11))#..................(2)

Using #d/(dx) (sqrt(x))=1/2(sqrt(x))# in equation (2) with chain rule
and also put g= 8t+11 in place of g in equation (2)

#= -sin(sqrt(8t+11))*1/(2(sqrt(8t+11)))*8#

On simplifying
#= -4sin(sqrt(8t+11))/((sqrt(8t+11)))#

Evelyn Agard · Answer

undefined To differentiate $ y = \cos(\sqrt{8t + 11}) $ using the chain rule, follow these steps:

1. Identify the outer function: $ \cos(u) $, where $ u = \sqrt{8t + 11} $.
2. Differentiate the outer function with respect to its inner variable: $ \frac{d}{du}\left[\cos(u)\right] = -\sin(u) $.
3. Identify the inner function: $ u = \sqrt{8t + 11} $.
4. Differentiate the inner function with respect to $ t $: $ \frac{du}{dt} = \frac{1}{2\sqrt{8t + 11}} $.
5. Apply the chain rule: $ \frac{dy}{dt} = -\sin(\sqrt{8t + 11}) \cdot \frac{1}{2\sqrt{8t + 11}} $.
6. Simplify the expression if necessary.