Find from first principles, the derivative of $\log (\sec (x^2))$
My Attempt :
Let, $y=f(x)=\log (\sec (x^2))$
$f(x+h)=\log (\sec (x+h)^2)$, where $h$ is a small increment in $x$
By first principle,
$$f'(x)=\lim_{h\to 0} \dfrac {f(x+h)-f(x)}{h}$$
$$=\lim_{h\to 0} \dfrac {\log (\sec (x+h)^2)-\log(\sec (x^2))}{h}$$
Best Answer
$$\ln(\sec(x+h^2))-\ln(\sec x^2)=\ln\dfrac{\cos x^2}{\cos(x+h)^2}$$
$$=\ln\left(1+\dfrac{\cos x^2-\cos(x+h)^2}{\cos(x+h)^2}\right)$$
Now $\lim_{h\to}\dfrac{\ln(1+h)}h=1$
So, we need to find $\lim_{h\to0}\dfrac{\dfrac{\cos x^2-\cos(x+h)^2}{\cos(x+h)^2}}h$ $$=\dfrac1{\lim_{h\to0}\cos(x+h)^2[\cos x^2+\cos(x+h)^2]}\cdot\lim_{h\to0}\dfrac{\sin^2(x+h)^2-\sin^2(x^2)}h$$
Using Prove $ \sin(A+B)\sin(A-B)=\sin^2A-\sin^2B $,
$\lim_{h\to0}\dfrac{\sin^2(x+h)^2-\sin^2(x^2)}h=\lim_{h\to0}\sin(x^2+(x+h)^2)\lim_{h\to0}\dfrac{\sin(2hx+h^2)}{2hx+h^2}\lim_{h\to0}\dfrac{2hx+h^2}h=?$