## Calculus: Early Transcendentals 8th Edition

(a) $$f'(x)=\frac{\sec^2x-\tan^2 x+\tan x}{\sec x}$$ (b) $$f'(x)=\cos x+\sin x$$ (c) Simplify the result in part (a). Change $\sec x$ into $\frac{1}{\cos x}$ and $\tan x$ into $\frac{\sin x}{\cos x}$
$$f(x)=\frac{\tan x-1}{\sec x}$$ (a) Use the Quotient Rule: $$f'(x)=\frac{(\tan x-1)'\sec x-(\tan x-1)(\sec x)'}{\sec^2 x}$$ $$f'(x)=\frac{\sec^2x\sec x-(\tan x-1)\sec x\tan x}{\sec^2 x}$$ $$f'(x)=\frac{\sec^2x-(\tan x-1)\tan x}{\sec x}$$ $$f'(x)=\frac{\sec^2x-\tan^2 x+\tan x}{\sec x}$$ (b) Simplify $f(x)$ $$f(x)=\frac{\tan x-1}{\sec x}$$ $$f(x)=\frac{\frac{\sin x}{\cos x}-1}{\frac{1}{\cos x}}$$ $$f(x)=\frac{\frac{\sin x-\cos x}{\cos x}}{\frac{1}{\cos x}}$$ $$f(x)=\frac{(\sin x-\cos x)\cos x}{\cos x\times 1}$$ $$f(x)=\sin x-\cos x$$ Therefore, $$f'(x)=\cos x-(-\sin x)=\cos x+\sin x$$ (c) Simplify the result in part (a) $$f'(x)=\frac{\sec^2x-\tan^2 x+\tan x}{\sec x}$$ $$f'(x)=\frac{\frac{1}{\cos^2 x}-\frac{\sin^2x }{\cos^2x}+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$$ $$f'(x)=\frac{\frac{1-\sin^2x}{\cos^2x}+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$$ $$f'(x)=\frac{\frac{\cos^2x}{\cos^2x}+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$$ $$f'(x)=\frac{1+\frac{\sin x}{\cos x}}{\frac{1}{\cos x}}$$ $$f'(x)=\frac{\frac{\sin x+\cos x}{\cos x}}{\frac{1}{\cos x}}$$ $$f'(x)=\frac{\sin x+\cos x}{1}=\sin x+\cos x$$ So 2 results from part a) and b) are equivalent.