Giải thích các bước giải:
Câu 1:
Ta có :
$f(x)=\dfrac{\cos^2x}{1+\sin^2x}$
$\to f'(x)=(\dfrac{\cos^2x}{1+\sin^2x})'$
$\to f'(x)=\dfrac{\left(\cos ^2\left(x\right)\right)'\left(1+\sin ^2\left(x\right)\right)-\left(1+\sin ^2\left(x\right)\right)'\cos ^2\left(x\right)}{\left(1+\sin ^2\left(x\right)\right)^2}$
$\to f'(x)=\dfrac{\left(-\sin \left(2x\right)\right)\left(1+\sin ^2\left(x\right)\right)-\sin \left(2x\right)\cos ^2\left(x\right)}{\left(1+\sin ^2\left(x\right)\right)^2}$
$\to f'(x)=-\dfrac{2\sin \left(2x\right)}{\left(1+\sin ^2\left(x\right)\right)^2}$
$\to f(\dfrac{\pi}{4})-3f'(\dfrac{\pi}{3})=\dfrac{\cos^2(\dfrac{\pi}{4})}{1+\sin^2(\dfrac{\pi}{4})}-3\cdot(-\dfrac{2\sin \left(\dfrac{2\pi}{3}\right)}{\left(1+\sin ^2\left(\dfrac{\pi}{3}\right)\right)^2})$
$\to f(\dfrac{\pi}{4})-3f'(\dfrac{\pi}{3})=\dfrac{49+144\sqrt{3}}{147}$
Câu 2:
Ta có :
$f(x)=\dfrac{1}{\sin^2x}$
$\to f'(x)=(\dfrac{1}{\sin^2x})'$
$\to f'(x)=\dfrac{0.\sin^2x-1.2\sin x.\cos x}{\sin^4x}$
$\to f'(x)=\dfrac{-\sin2x}{\sin^4x}$
