How do you use the chain rule to differentiate #y=cos(sqrt(8t+11))#?
Risposta:
La risposta è #-4sin(sqrt(8t+11))/((sqrt(8t+11)))#
Spiegazione:
Differentiating y with respect to t.
#dy/dt=d/dt (cos(sqrt8t+11)) #
utilizzando Regola di derivazione
#(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)
utilizzando #d/(dx)(cos (x))=-sin(x)# solve the equation (1)
#= -sin(g)*d/(dt)(sqrt(8t+11))#..................(2)
utilizzando #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)))#