find the derivative of each function.\n1.) $f(\\theta)=\frac{\\pi}{2}\\sin\\theta - \\cos\\theta$\n2.) $y =…

find the derivative of each function.\n1.) $f(\\theta)=\frac{\\pi}{2}\\sin\\theta - \\cos\\theta$\n2.) $y = x^{2}-\frac{1}{2}\\cos x$\n3.) $f(x)=\frac{1}{2}e^{x}-3\\sin x$\n4.) $f(x)=\frac{1}{x^{2}}-2e^{x}$

find the derivative of each function.\n1.) $f(\\theta)=\frac{\\pi}{2}\\sin\\theta - \\cos\\theta$\n2.) $y = x^{2}-\frac{1}{2}\\cos x$\n3.) $f(x)=\frac{1}{2}e^{x}-3\\sin x$\n4.) $f(x)=\frac{1}{x^{2}}-2e^{x}$

Answer

Explanation:

Step1: Recall derivative rules

Recall that $\frac{d}{dx}(\sin x)=\cos x$, $\frac{d}{dx}(\cos x)=-\sin x$, $\frac{d}{dx}(e^{x}) = e^{x}$ and $\frac{d}{dx}(x^{n})=nx^{n - 1}$.

Step2: Differentiate $f(\theta)=\frac{\pi}{2}\sin\theta-\cos\theta$

Using the sum - difference rule of derivatives $\frac{d}{d\theta}(u - v)=\frac{d u}{d\theta}-\frac{d v}{d\theta}$ and the constant - multiple rule $\frac{d}{d\theta}(cu)=c\frac{d u}{d\theta}$ ($c$ is a constant), we have $f'(\theta)=\frac{\pi}{2}\cos\theta-(-\sin\theta)=\frac{\pi}{2}\cos\theta+\sin\theta$.

Step3: Differentiate $y = x^{2}-\frac{1}{2}\cos x$

Applying the sum - difference rule and constant - multiple rule, $y'=\frac{d}{dx}(x^{2})-\frac{1}{2}\frac{d}{dx}(\cos x)=2x+\frac{1}{2}\sin x$.

Step4: Differentiate $f(x)=\frac{1}{2}e^{x}-3\sin x$

Using the sum - difference and constant - multiple rules, $f'(x)=\frac{1}{2}\frac{d}{dx}(e^{x})-3\frac{d}{dx}(\sin x)=\frac{1}{2}e^{x}-3\cos x$.

Step5: Differentiate $f(x)=\frac{1}{x^{2}}-2e^{x}$

Rewrite $\frac{1}{x^{2}}=x^{- 2}$. Then, by the sum - difference and constant - multiple rules, $f'(x)=\frac{d}{dx}(x^{-2})-2\frac{d}{dx}(e^{x})=-2x^{-3}-2e^{x}=-\frac{2}{x^{3}}-2e^{x}$.

Answer:

  1. $f'(\theta)=\frac{\pi}{2}\cos\theta+\sin\theta$
  2. $y'=2x+\frac{1}{2}\sin x$
  3. $f'(x)=\frac{1}{2}e^{x}-3\cos x$
  4. $f'(x)=-\frac{2}{x^{3}}-2e^{x}$