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find f'(x) f(x)=1/(secx)^7
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\[\Large\bf\sf \sec x\quad=\quad \frac{1}{\cos x}\] Use that^\[\Large\bf\sf \left(\frac{1}{\sec x}\right)^7\quad=\quad ?\]
\[\frac{ 1 }{ (secx)^7 }\]
??
Use the identity.
\[\Large\bf\sf \sec x\quad=\quad \frac{1}{\cos x}\qquad\implies\qquad \frac{1}{\sec x}\quad=\quad \cos x\]
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\[\Large\bf\sf \left(\frac{1}{\sec x}\right)^7\quad=\quad ?\]
so it wont be secx=secxtanx
cos (x)
\[\Large\bf\sf \left(\frac{1}{\sec x}\right)^7\quad=\quad (\cos x)^7\]
Going off of what zepdrix said, you then use the Chain-Chain Rule to find f'(x) = 7(cosx) * - sinx => f'(x) = -7sinxcosx.
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*7(cosx)^6 * -sinx => f'(x) = -7(sinx)(cosx)^6
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