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Question
find the slope of the graph of the function at the given point. use proper notation.
5.) $f(\theta)=4\sin\theta - \theta,\\ (0,0)$
6.) $f(x)=\frac{3}{4}e^{x},\\ (0,\frac{3}{4})$
Step1: Differentiate the function $f(\theta)$
The derivative of $\sin\theta$ is $\cos\theta$ and the derivative of $\theta$ is $1$. Using the sum - difference rule of differentiation, if $f(\theta)=4\sin\theta-\theta$, then $f'(\theta)=\frac{d}{d\theta}(4\sin\theta)-\frac{d}{d\theta}(\theta)$. So $f'(\theta)=4\cos\theta - 1$.
Step2: Evaluate the derivative at the given point
We want to find the slope at the point $(0,0)$. Substitute $\theta = 0$ into $f'(\theta)$. Since $\cos(0)=1$, then $f'(0)=4\cos(0)-1=4\times1 - 1=3$.
for second function:
Step1: Differentiate the function $f(x)$
The derivative of $e^{x}$ is $e^{x}$. If $f(x)=\frac{3}{4}e^{x}$, then by the constant - multiple rule of differentiation $f'(x)=\frac{3}{4}\frac{d}{dx}(e^{x})$. So $f'(x)=\frac{3}{4}e^{x}$.
Step2: Evaluate the derivative at the given point
We want to find the slope at the point $(0,\frac{3}{4})$. Substitute $x = 0$ into $f'(x)$. Since $e^{0}=1$, then $f'(0)=\frac{3}{4}e^{0}=\frac{3}{4}\times1=\frac{3}{4}$.
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The slope of the graph of the function $f(\theta)=4\sin\theta-\theta$ at the point $(0,0)$ is $3$.