Let y=tan(3x+4) Find the differential dy when x=5 and dx=0.3 Find the differential dy when x=5 and dx=0.6

Problem 1

y = tan(3x+4)

f(x) = tan(3x+4)

f ' (x) = 3sec^2(3x+4) .... apply derivative chain rule

dy/dx = f ' (x)

dy = f ' (x) * dx

dy = ( 3sec^2(3x+4) ) * dx

Now plug in x = 5 and dx = 0.3

dy = ( 3sec^2(3*5+4) ) * 0.3

dy = 0.920681 which is approximate

Make sure your calculator is in radian mode.  Calculus textbooks will be in radian mode for the special sine limit definition to be true.

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Problem 2

We'll have the same derivative function and same x value. The only difference is that dx = 0.6 this time.

dy = f ' (x) * dx

dy = ( 3sec^2(3*5+4) ) * 0.6

dy = 1.84136 approximately

yes

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