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 [tex]\lim_{x\to0}\frac{\sin x}{x} = 1[/tex] 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
