calculate the derivative of the following function.\ny = cos^9θ+ sin^8θ\n\frac{dy}{dθ}=□

Explanation:

Step1: Apply sum - rule of derivatives

The sum - rule states that if $y = u + v$, then $\frac{dy}{d\theta}=\frac{du}{d\theta}+\frac{dv}{d\theta}$. Let $u = \cos^{9}\theta$ and $v=\sin^{8}\theta$. So, $\frac{dy}{d\theta}=\frac{d(\cos^{9}\theta)}{d\theta}+\frac{d(\sin^{8}\theta)}{d\theta}$.

Step2: Use chain - rule for $\frac{d(\cos^{9}\theta)}{d\theta}$

The chain - rule is $\frac{d(f(g(\theta)))}{d\theta}=f^\prime(g(\theta))\cdot g^\prime(\theta)$. Let $y = u^{9}$ and $u=\cos\theta$. Then $\frac{dy}{du}=9u^{8}$ and $\frac{du}{d\theta}=-\sin\theta$. So, $\frac{d(\cos^{9}\theta)}{d\theta}=9\cos^{8}\theta\cdot(-\sin\theta)=- 9\cos^{8}\theta\sin\theta$.

Step3: Use chain - rule for $\frac{d(\sin^{8}\theta)}{d\theta}$

Let $y = u^{8}$ and $u = \sin\theta$. Then $\frac{dy}{du}=8u^{7}$ and $\frac{du}{d\theta}=\cos\theta$. So, $\frac{d(\sin^{8}\theta)}{d\theta}=8\sin^{7}\theta\cdot\cos\theta$.

Step4: Combine the results

$\frac{dy}{d\theta}=-9\cos^{8}\theta\sin\theta + 8\sin^{7}\theta\cos\theta$.

Answer:

$-9\cos^{8}\theta\sin\theta + 8\sin^{7}\theta\cos\theta$