I have to find out the derivative of given function by using 'first principle method', but the question has eaten half of my brain, the problem is first principle method. The function is $f(x)= \tan x\sec x$.
[Math] The derivative of $f(x) = \tan x\sec x$ using first principles
calculusderivatives
Best Answer
Because they are more familiar, we express $f(x)$ in terms of sines and cosines. We get $$f(x)=\frac{\sin x}{\cos^2 x}=\frac{\sin x}{1-\sin^2 x}.$$ Thus $$f'(x)=\lim_{h\to 0} \frac{\frac{\sin(x+h)}{1-\sin^2(x+h)}-\frac{\sin x}{1-\sin^2 x}}{h}.$$ We will need to impose the condition $\sin x\ne \pm 1$. Bring the top to the common denominator $(1-\sin^2(x+h))(1-\sin^2 x)$. Some algebra shows that we want $$\lim_{h\to 0}\frac{1+\sin(x+h)\sin x}{(1-\sin^2(x+h))(1-\sin^2 x)} \frac{\sin(x+h)-\sin x}{h}.$$
Now copy the standard first principles argument that shows that $\lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}=\cos x$. The limits of the remaining terms are easy to compute.