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G(t)=t^2e^-t + ((ln(t))^2)

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Find the derivative of the following via implicit differentiation:d/dt(G(t)) = d/dt(t^2/e^t + log^2(t))
The derivative of G(t) is G'(t):G'(t) = d/dt(t^2/e^t + log^2(t))
Differentiate the sum term by term:G'(t) = d/dt(t^2/e^t) + d/dt(log^2(t))
Use the product rule, d/dt(u v) = v ( du)/( dt) + u ( dv)/( dt), where u = e^(-t) and v = t^2:G'(t) = d/dt(log^2(t)) + t^2 d/dt(e^(-t)) + (d/dt(t^2))/e^t
Simplify the expression:G'(t) = t^2 (d/dt(e^(-t))) + (d/dt(t^2))/e^t + d/dt(log^2(t))
Using the chain rule, d/dt(e^(-t)) = ( d e^u)/( du) ( du)/( dt), where u = -t and ( d)/( du)(e^u) = e^u:G'(t) = (d/dt(t^2))/e^t + d/dt(log^2(t)) + (d/dt(-t))/e^t t^2
Factor out constants:G'(t) = (d/dt(t^2))/e^t + d/dt(log^2(t)) + (-d/dt(t) t^2)/e^t
The derivative of t is 1:G'(t) = (d/dt(t^2))/e^t + d/dt(log^2(t)) - (1 t^2)/e^t
Use the power rule, d/dt(t^n) = n t^(n - 1), where n = 2: d/dt(t^2) = 2 t:G'(t) = -t^2/e^t + d/dt(log^2(t)) + (2 t)/e^t
Using the chain rule, d/dt(log^2(t)) = ( du^2)/( du) ( du)/( dt), where u = log(t) and ( d)/( du)(u^2) = 2 u:G'(t) = (2 t)/e^t - t^2/e^t + 2 log(t) d/dt(log(t))
The derivative of log(t) is 1/t:Answer: G'(t) = (2 t)/e^t - t^2/e^t + 1/t 2 log(t)
User Danila Alpatov
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