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E2_4A4

Example 4 Assume that P(5)=7, Q(–3)=5, P'(5)=6, Q'(–3)= –4. Find the equation of the tangent line T to the graph of y = (PoQ)(x) at x= –3 . Solution We use the chain rule to compute the slope ( PoQ ) '(–3) of this tangent line:

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E2_4A4

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  1. Example 4 Assume that P(5)=7, Q(–3)=5, P'(5)=6, Q'(–3)= –4. Find the equation of the tangent line T to the graph of y =(PoQ)(x) at x= –3. Solution We use the chain rule to compute the slope (PoQ)'(–3) of this tangent line: The point (–3,(PoQ)(–3)) = (–3,P(Q(–3))) = (–3,P(5)) = (–3,7) lies on the tangent line T. By the point-slope formula the line T through the point (–3,7)with slope –24 has equation: y = 7 – 24[x – (–3)] = 7 – 24x – 72 = –65 – 24x

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