The derivative of \( \sec^5\left(\frac{ax + b}{c}\right) \) with respect to \( x \) is \(\frac{5a}{c} \sec^4\left(\frac{ax + b}{c}\right) \cdot \sec\left(\frac{ax + b}{c}\right) \tan\left(\frac{ax + b}{c}\right)\).
To find the derivative of \( \sec^5\left(\frac{ax + b}{c}\right) \) with respect to \( x \), you can use the chain rule. The chain rule states that if you have a composite function \( g(f(x)) \), then the derivative is \( g'(f(x)) \cdot f'(x) \).
Let \( u = \frac{ax + b}{c} \). Then the given function can be written as \( \sec^5(u) \).
Now, differentiate with respect to \( x \):
\[ \frac{d}{dx} \sec^5(u) = 5 \sec^4(u) \cdot \sec(u) \tan(u) \cdot \frac{du}{dx} \]
Now, substitute back \( u = \frac{ax + b}{c} \) and multiply by \( \frac{du}{dx} \):
\[ 5 \sec^4\left(\frac{ax + b}{c}\right) \cdot \sec\left(\frac{ax + b}{c}\right) \tan\left(\frac{ax + b}{c}\right) \cdot \frac{a}{c} \]
So, the derivative is:
\[ \frac{5a}{c} \sec^4\left(\frac{ax + b}{c}\right) \cdot \sec\left(\frac{ax + b}{c}\right) \tan\left(\frac{ax + b}{c}\right) \]