i. $\frac{d}{dx}(4cos^{7}(x)-10sec^{2}(x))$

Explanation:

Step1: Apply sum - difference rule

$\frac{d}{dx}(4\cos^{7}(x)- 10\sec^{2}(x))=\frac{d}{dx}(4\cos^{7}(x))-\frac{d}{dx}(10\sec^{2}(x))$

Step2: Factor out constants

$4\frac{d}{dx}(\cos^{7}(x)) - 10\frac{d}{dx}(\sec^{2}(x))$

Step3: Use chain - rule for $\frac{d}{dx}(\cos^{7}(x))$

Let $u = \cos(x)$, then $y = u^{7}$. $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. $\frac{dy}{du}=7u^{6}=7\cos^{6}(x)$ and $\frac{du}{dx}=-\sin(x)$. So $\frac{d}{dx}(\cos^{7}(x)) = 7\cos^{6}(x)\cdot(-\sin(x))=-7\cos^{6}(x)\sin(x)$

Step4: Use chain - rule for $\frac{d}{dx}(\sec^{2}(x))$

Let $u=\sec(x)$, then $y = u^{2}$. $\frac{dy}{du}=2u = 2\sec(x)$ and $\frac{du}{dx}=\sec(x)\tan(x)$. So $\frac{d}{dx}(\sec^{2}(x))=2\sec(x)\cdot\sec(x)\tan(x)=2\sec^{2}(x)\tan(x)$

Step5: Substitute back into the expression

$4\times(-7\cos^{6}(x)\sin(x))-10\times(2\sec^{2}(x)\tan(x))=-28\cos^{6}(x)\sin(x)-20\sec^{2}(x)\tan(x)$

Answer:

$-28\cos^{6}(x)\sin(x)-20\sec^{2}(x)\tan(x)$