If $\displaystyle f(u) = \sin{u}$ and $\displaystyle u = g(x) = x^2-9$, then $\displaystyle (f \circ g)'(3) = $ I know that $\displaystyle g'(x)= 2x $ but didn't know how to use this, the answer is 6
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you got this : f composition g = sin(x^2-9) finding it's derivative you get this .. cos(x^2-9)*2x ,substituting 3 yields cos(9-9)*2*3=1*6=6 gl
Originally Posted by bigwave If $\displaystyle f(u) = \sin{u}$ and $\displaystyle u = g(x) = x^2-9$, then $\displaystyle (f \circ g)'(3) = $ I know that $\displaystyle g'(x)= 2x $ but didn't know how to use this, the answer is 6 So, $\displaystyle (f \circ g)=\sin{(x^{2}-9)}$ $\displaystyle (f \circ g)' = \cos{(x^{2}-9)} \times 2x$
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