# Video: Finding the Derivative of a Vector-Valued Function

Find the derivative of the vector-valued function π(π‘) = π^(π‘) π + π^(βπ‘) π + 3π.

03:12

### 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 πth component of the derivative of a π-dimensional vector-valued function π of π‘ with respect to π‘ is just the derivative of the πth 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 ππ‘ with respect to π‘, where π is a constant, is π multiplied by π to the power of ππ‘. 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 donβt 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.