Questions: The definition of the derivative allows us to define a tangent line precisely. Definition. Let f be a function differentiable at a. The line tangent to the curve y=f(x) at the point (a, f(a)) is the line that passes through the point (a, f(a)) whose slope is equal to f'(a). Naturally, by the point-slope equation of the line, it follows that the tangent line is given by the equation y=f(a)+f'(a)(x-a) Question. Let f be a function given by f(x)=4x-3. What is the instantaneous rate of change of f at a, where a is any real number? The rate of change is given by f'(a), and so is the slope of the tangent line to the curve y=f(x) at the point (a, f(a)). The derivative is the slope of the line y=4x-3 ! Therefore, f'(a)= for all real numbers a. Example. Let f(x)=2x^2+3. Find the slope of the tangent line to the curve y=f(x) at the point (2, f(2)). Explanation. Finding the slope of the tangent line at the point (2, f(2)) means finding f'(2) f'(2)=lim (x -> 2) (f(x)-f(2))/(x-2) Therefore, the slope of the tangent line is f'(2)=lim (x -> 2) (f(x)-f(2))/(x-2)=8

Transcript text: The definition of the derivative allows us to define a tangent line precisely. Definition. Let $f$ be a function differentiable at $a$. The line tangent to the curve $y=f(x)$ at the point $(a, f(a))$ is the line that passes through the point $(a, f(a))$ whose slope is equal to $f^{\prime}(a)$. Naturally, by the point-slope equation of the line, it follows that the tangent line is given by the equation \[ y=f(a)+f^{\prime}(a)(x-a) \] Question. Let $f$ be a function given by $f(x)=4 x-3$. What is the instantaneous rate of change of $f$ at $a$, where $a$ is any real number? The rate of change is given by $f^{\prime}(a)$, and so is the slope of the tangent line to the curve $y=f(x)$ at the point $(a, f(a))$. The derivative is the slope of the line $y=4 x-3$ ! Therefore, $f^{\prime}(a)=$ $\square$ for all real numbers $a$. Example. Let $f(x)=2 x^{2}+3$. Find the slope of the tangent line to the curve $y=f(x)$ at the point $(2, f(2))$. Explanation. Finding the slope of the tangent line at the point $(2, f(2))$ means finding $f^{\prime}(2)$ \[ f^{\prime}(2)=\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2} \] Therefore, the slope of the tangent line is \[ f^{\prime}(2)=\lim _{x \rightarrow 2} \frac{f(x)-f(2)}{x-2}=8 \]