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Find f’(2),where f(t)=u(t).v(t), \(u(2) = \left\langle {1,2, - 1} \right\rangle \)and \(u'(2) = \left\langle {3,0,4} \right\rangle \) \(v(t) = \left\langle {t,{t^2},{t^3}} \right\rangle \)

Short Answer

Expert verified

\(f'(2) = 35\)

Step by step solution

01

Step 1: Differentiating component wise

We have \(v(t) = (t,{t^2},{t^3})\)

\(\begin{array}{l}v'(t) = (1,2t,3{t^2})\\t = 2\\v(2) = (2,4,8)\end{array}\)

\(\begin{array}{l}u'(2) = (3,0,4)\\f(t) = v(t).u(t)\\f'(t) = v'(t).u(t) + v(t).u'(t)\\f'(2) = v'(2).u(2) + v(2).u'(2)\end{array}\)

\(\)

02

Step 2: Substitution

On substitution we get

\(\begin{array}{l}f'(2) = (1,4,12) \cdot (1,2, - 1) + (2,4,8) \cdot (3,0,4) = 35\\f'(2) = 35\end{array}\)

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