Question Video: Derivatives of Vector-Valued Functions | Nagwa Question Video: Derivatives of Vector-Valued Functions | Nagwa

# Question Video: Derivatives of Vector-Valued Functions Mathematics • Higher Education

Find the derivative of the vector-valued function π«(π‘) = (β(3π‘))π’ β (π^(9π‘))π£.

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### Video Transcript

Find the derivative of the vector-valued function π« of π‘ equals a square root of three π‘ π’ minus π to the power of nine π‘ π£.

In this question, we are asked to differentiate the given vector-valued function π« of π‘ with respect to the variable π‘. Recall that the derivative of a vector-valued function is obtained by differentiating each of its component functions. Therefore, in order to find the derivative of the vector-valued function π« of π‘ given to us in the question, we need to differentiate the component functions square root of three π‘ and π to the power of nine π‘ with respect to π‘.

In order to find the derivative of the function square root of three π‘ with respect to π‘, note that we can rewrite the function as a square root of three multiplied by square root of π‘. Further, letβs use the fact that the square root of a number or variable can be written as that number or variable raised to the power of a half. Using this, we can rewrite square root of π‘ as π‘ to the power of a half. Having done so, we can now use the power rule for differentiation to differentiate square root of three π‘ with respect to π‘.

Multiplying the coefficient, square root of three, by the exponent or power, one-half, and then decreasing the exponent by one, we obtain that the derivative of square root of three π‘ with respect to π‘ is square root of three timesed by a half timesed by a π‘ to the power of negative a half. We can rewrite this as square root of three over two multiplied by one over π‘ to the power of a half and further as a square root of three over two times square root of π‘. We can keep this expression as it is or multiply it by square root of three over square root of three, which equals one, in order to obtain three over two times square root of three π‘.

Having found the derivative of the function square root of three π‘ with respect to π‘, letβs now find the derivative of the function π to the power of nine π‘ with respect to π‘. We have that for any constant π, the derivative of the function π to the power of ππ₯ with respect to π₯ is ππ to the power of ππ₯. Therefore, the derivative of the function π to the power of nine π‘ with respect to π‘ is nine π to the power of nine π‘.

So, the derivative with respect to π‘ of the vector-valued function π« of π‘ equals a square root of three π‘ π’ minus π to the power of nine π‘ π£ is three over two times square root of three π‘ π’ minus nine times π to the power of of nine π‘ π£. This is our final answer.

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