# Question Video: Finding the Rate of Change of Exponential Functions in a Real-World Context Mathematics

A factoryβs production of π¦ units in day π‘ is given by the relation π¦ = 400 (10 β π^(β0.8π‘). What is the rate of change of production with respect to time on the fifth day?

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### Video Transcript

A factoryβs production of π¦ units in day π‘ is given by the relation π¦ is equal to 400 multiplied by 10 minus π to the power of negative 0.8π‘. What is the rate of change of production with respect to time on the fifth day?

The question tells us that a factoryβs production of π¦ units on day π‘ is given by π¦ is equal to 400 multiplied by 10 minus π to the power of negative 0.8π‘. And it wants us to find the rate of change of production with respect to time on the fifth day.

We recall the rate of change of production with respect to time is the same as differentiating our expression for production with respect to time. And the question wants us evaluate this on the fifth day. Thatβs when π‘ is equal to five. So the question is asking us to differentiate 400 multiplied by 10 minus π to the power of negative 0.8π‘ with respect to π‘.

We know, for any constant π, the derivative of π times a function π of π‘ with respect to π‘ is equal to π times the derivative of π of π‘ with respect to π‘. We can take a constant factor outside of our derivative. So we can take the constant factor of 400 outside of our derivative. This gives us 400 multiplied by the derivative of 10 minus π to the power of negative 0.8π‘ with respect to π‘.

Next, weβre going to use the fact that the derivative of a sum or difference of two functions is equal to the sum or difference of their derivatives. This gives us 400 multiplied by the derivative of 10 with respect to π‘ minus the derivative of π to the power of negative 0.8π‘ with respect to π‘.

Weβre now ready to evaluate this expression. We know the derivative of any constant is equal to zero. So the derivative of 10 with respect to π‘ is equal to zero. And we also know, for any constant π, the derivative of π raised to the power of ππ‘ with respect to π‘ is just equal to π multiplied by π to the power of ππ‘. We multiplied by the coefficient in the exponent.

So the derivative of π to the power of negative 0.8π‘ with respect to π‘ is equal to negative 0.8π to the power of negative 0.8π‘. And since we were subtracting our derivative, this gives us 400 multiplied by 0.8 times π to the power of negative 0.8π‘. Finally, we have 400 multiplied by 0.8 is equal to 320.

Therefore, we found an expression for the rate of change of production in our factory with respect to time. And the question wants us to find the rate of change of production on the fifth day. So we substitute π‘ is equal to five into this expression. Substituting π‘ is equal to five gives us the rate of change of production in this factory on the fifth day is equal to 320π to the power of negative 0.8 times five. And we can simplify our exponent of negative 0.8 times five to be negative four.

So by taking our negative exponent into the denominator of our fraction. Weβve shown that the rate of change of production for this factory with respect to time on the fifth day is equal to 320 divided by π to the fourth power.