Chain Quotient Order

It's actually a product rule question in disguise.

We can split $f(x) = \frac{e^{\sin x}}{\sec^{2} x} = e^{\sin x}\cos^{2} x$ into $g(x) = e^{\sin x}$ and $h(x) = \cos^{2} x$ .

Then, $f'(x) = g(x)h'(x) + g'(x)h(x)$ by the product rule. This means we have to find $g'(x)$ and $h'(x)$ first. By the chain rule,

$g'(x) = \left( e^{\sin x} \right)' = e^{\sin x}\cos x$

$h'(x) = \left( \cos^{2} x \right)' = -2\cos x \sin x$

Hence,

$f'(x) = e^{\sin x}\cos^{3} x - 2 e^{\sin x}\cos x \sin x$

We can then substitute in $x = 0$ to obtain $f'(0) = \boxed{1}$ .

Applying y=u/v form so here u = e^sinx and v = sec^2x Derivative of u= cosxe^sinx and v = 2secx secxtanx General derivative rule of u/v is (U dv/dx -vdv/dx)/v^2 f'(x)= cosxe^sinx - 2 secx secx tandem e^sinx / sec^4x f'(0)= 1 Because e^0=1, sec0=1, tan0= 0