Pregunta
Find the derivative of the function. \[ f(t)=88^{t^{3}} \] \( f^{\prime}(t)=\square \)
Ask by Bolton Griffiths. in the United States
Mar 15,2025
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The derivative of \( f(t) = 88^{t^3} \) is \( f'(t) = 3t^2 \ln(88) \cdot 88^{t^3} \).
Solución
We start with the function
\[
f(t)=88^{t^3}.
\]
Recall that if we have a function of the form
\[
g(t)=a^{h(t)},
\]
its derivative is given by
\[
g'(t)=a^{h(t)}\ln(a)\cdot h'(t).
\]
For our function, \( a=88 \) and \( h(t)=t^3 \). Now, we compute the derivative of \( h(t) \):
\[
h'(t)=\frac{d}{dt}(t^3)=3t^2.
\]
Thus, applying the formula we get:
\[
f'(t)=88^{t^3}\ln(88)\cdot 3t^2.
\]
Rearranging the factors, the derivative is:
\[
f'(t)=3t^2\ln(88)\cdot88^{t^3}.
\]
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Simplifique esta solución Extra Insights
To find the derivative of the function \( f(t) = 88^{t^3} \), we can use the chain rule and the exponential rule. The derivative of \( a^u \) is given by \( a^u \ln(a) \cdot u' \). Here, \( u = t^3 \) and \( u' = 3t^2 \). So, we have: \[ f'(t) = 88^{t^3} \ln(88) \cdot 3t^2 = 3t^2 \cdot 88^{t^3} \ln(88). \] Thus, the derivative is: \[ f'(t) = 3t^2 \cdot 88^{t^3} \ln(88). \] Happy differentiating!
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