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

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Answer 1 · 2018-02-21T22:02:51

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

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

#(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)))#