### Video Transcript

Find the derivative of the vector-valued function π of π‘ equals π to the power of π‘ π plus π to the power of negative π‘ π plus three π.

Recall that the derivative with respect to π‘ of a three-dimensional vector-valued function π of π‘ equal to π of π‘ π plus π of π‘ π plus β of π‘ π is just π prime of π‘ equal to π prime of π‘ π plus π prime of π‘ π plus β prime of π‘ π. More generally, the πth component of the derivative of a π-dimensional vector-valued function π of π‘ with respect to π‘ is just the derivative of the πth component of π with respect to π‘.

So the first component of the derivative of the three-dimensional vector-valued function π given to us in the question is equal to the derivative of the function π to the power of π‘ with respect to π‘. The second component of the derivative is equal to the derivative of the function π to the power of negative π‘ with respect to π‘. The third component of the derivative is equal to the derivative of the constant function three with respect to π‘.

Now recall that the derivative of the exponential function π to the power of π‘ with respect to π‘ is just π to the power of π‘, i.e., itself. More generally, the derivative of the function π to the power of ππ‘ with respect to π‘, where π is a constant, is π multiplied by π to the power of ππ‘. If we let π equal negative one in this formula, then we obtain that the derivative of π to the power of negative π‘ with respect to π‘ is equal to negative π to the power of negative π‘. Therefore, the first two components of the derivative with respect to π‘ of the vector-valued function π given to us in the question are π to the power of π‘ and negative π to the power of negative π‘, respectively.

Now recall that the derivative of a constant is zero with respect to any variable. Therefore, the third component of the derivative with respect to π‘ of the vector-valued function π given to us in the question is zero. We donβt need to write down the third component, zero, explicitly. So our final answer is that the derivative with respect to π‘ of the vector-valued function π of π‘ given to us in the question is π prime of π‘ equals π to the power of π‘ π minus π to the power of negative π‘ π.