# Video: Evaluating a Limit Using the Definition of Differentiation

Calculate πβ²(π ), and find the vector form of the equation of the tangent line at π(0) for π(π ) = (π  + 1, π Β² + 1, π Β³ + 1).

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### Video Transcript

Calculate π prime of π  and find the vector form of the equation of the tangent line at π of zero for π of π  equals π  plus one, π  squared plus one, π  cubed plus one.

In this question, weβve been given a rule for π in terms of its π₯-, π¦-, and π§-coordinates. We can alternatively think of this as a vector-valued function such that the vector π of π  is equal to π  plus one, π  squared plus one, π  cubed plus one, where π  plus one is the π’-component, π  squared plus one is the π£-component, and π  cubed plus one is the π€-component.

Remember then, to find the derivative of a vector-valued function, we simply take the derivative of each component. This means weβre going to individually differentiate with respect to π , π  plus one, π  squared plus one, and π  cubed plus one. That gives us one, two π , and three π  squared, respectively. Taking this back into the rule for the π₯-, π¦-, and π§-coordinates, and we find π prime of π  to be equal to one, two s, three π  squared.

We do still need to find the vector form of the equation of the tangent line at π of zero. So, we recall that a line that passes through a point π₯ nought, π¦ nought, π§ nought with the direction vector π― is given by the equation π« equals π« nought plus π‘ times π―. So, weβll need two things, a point that our line passes through and its direction of travel. We can find the latter by evaluating the derivative when the parameter value, π , is equal to zero. So, letβs substitute π  equals zero into our derivative. Thatβs one, zero, zero or one π’ plus zero π£ plus zero π€.

We can also find that the point our tangent line passes through by substituting π  equals zero into the original function. Thatβs π of zero. When we do, we obtain π of zero to be zero plus one, zero plus one, zero plus one. Thatβs one, one, one. And so, the vector form of the equation of the tangent line at π zero, letβs call that πΏ, is one, one, one plus π‘ times one, zero, zero.