Suppose f(x) satisfies xcos(f(x))−f(x)= x^3 +8 and f(3)=0. Find f ’(3).?
Suppose f(x) satisfies xcos(f(x))−f(x)= x^3 +8 and f(3)=0. Find f ’(3).
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Suppose f(x) satisfies xcos(f(x))−f(x)= x^3 +8 and f(3)=0. Find f ’(3).
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Using the Product and Chain Rules, we have:
(d/dx){x*cos[f(x)] - f(x)} = (d/dx)(x^3 + 8)
==> cos[f(x)] - x*f'(x)*sin[f(x)] - f'(x) = 3x^2.
At x = 3, we have:
cos[f(3)] - 3*f'(3)*sin[f(3)] - f'(3) = 3(3)^2
==> cos(0) - 3*f'(3)*sin(0) - f'(3) = 27, since f(3) = 0
==> 1 - f'(3) = 27
==> f'(3) = -26.
I hope this helps!