derivative of y(t)=t^2-4ln(t+1)-1?

Answer 1

y'(t) = 2t - 4* derivative of ln(t+1) - 0

You must know that derivative of ln t is 1/t,
and since (t+1) has derivative 1 we can treat it as if it were t, so we get
1/(t+1),
which gives
y'(t) = 2t - 4/(t+1)

Answer 2

6

Answer 3

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Derivative of y(t) = dy/dt = d/dt ( t^2 - 4 ln(t+1) - 1 )

= 2t - 4 / t+1

This using the basic derivative formulae:

d/dx ( x^n ) = n * x^(n-1)
d/dx ( ln (ax+b) ) = a / ax + b
d/dx ( constant ) = 0

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Answer 4

y(t)=t^2-4ln(t+1)-1

y'(t) = d/dt(t^2-4ln(t+1)-1)

y'(t) = d/dt(t^2) - d/dt(4ln(t+1)) - d/dt(1)

Formulae to be used:
d/dt (t^n) = n * t^(n-1)
d/dt (ln (at+b)) = a / at + b
d/dt (constant) = 0


d/dt(t^2) = 2*t

d/dt(4ln(t+1)) = 4/(t+1)

d/dt(1) = 0

y'(t) = 2t - [4/(t+1)]

Answer 5

(2t^2 +2t -4)/(t+1)