2.3. Tutorial questions - Slopes of curves D) From first principle= (ie from the definition of the derivative) find \( \begin{array}{lll}\text { (a) } \frac{d y}{d f} \text { if } y=x^{2}+x+1 & \text { (b) } y \text { if } y=\frac{1}{x} & \text { (c) } f^{\prime}(x) \text { if } f(x)=\sqrt{x^{2}+1} \\ \text { (d) } \frac{d(f)}{d t} \text { if } f(t)=\frac{1}{2+\sqrt{t}}\end{array} \)

(a) For \( y=x^2+x+1 \): - The definition of the derivative is \[ f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}. \] - Here, \( f(x)=x^2+x+1 \) so that \[ f(x+h)=(x+h)^2+(x+h)+1=x^2+2xh+h^2+x+h+1. \] - Then, \[ f(x+h)-f(x)=(x^2+2xh+h^2+x+h+1)-(x^2+x+1)=2xh+h^2+h. \] - Dividing by \( h \), \[ \frac{f(x+h)-f(x)}{h}=2x+h+1. \] - Taking the limit \( h\to 0 \), \[ \frac{dy}{dx}=2x+1. \] --- (b) For \( y=\frac{1}{x} \): - Using the definition with \( f(x)=\frac{1}{x} \) and \( f(x+h)=\frac{1}{x+h} \), \[ f(x+h)-f(x)=\frac{1}{x+h}-\frac{1}{x}=\frac{x-(x+h)}{x(x+h)}=\frac{-h}{x(x+h)}. \] - Thus, \[ \frac{f(x+h)-f(x)}{h}=\frac{-h}{h\,x(x+h)}=\frac{-1}{x(x+h)}. \] - Taking the limit \( h\to 0 \), \[ \frac{dy}{dx}=-\frac{1}{x^2}. \] --- (c) For \( f(x)=\sqrt{x^2+1} \): - We have \[ f(x+h)=\sqrt{(x+h)^2+1}=\sqrt{x^2+2xh+h^2+1}. \] - Then the difference quotient is \[ \frac{f(x+h)-f(x)}{h}=\frac{\sqrt{x^2+2xh+h^2+1}-\sqrt{x^2+1}}{h}. \] - Multiply the numerator and the denominator by the conjugate \(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\): \[ \frac{\sqrt{x^2+2xh+h^2+1}-\sqrt{x^2+1}}{h} \cdot \frac{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}. \] - The numerator becomes \[ (x^2+2xh+h^2+1)-(x^2+1)=2xh+h^2. \] - Therefore, \[ \frac{2xh+h^2}{h\Bigl(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\Bigr)}=\frac{h(2x+h)}{h\Bigl(\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}\Bigr)}. \] - Cancel \( h \): \[ \frac{2x+h}{\sqrt{x^2+2xh+h^2+1}+\sqrt{x^2+1}}. \] - Taking \( h\to 0 \) gives \[ f'(x)=\frac{2x}{2\sqrt{x^2+1}}=\frac{x}{\sqrt{x^2+1}}. \] --- (d) For \( f(t)=\frac{1}{2+\sqrt{t}} \): - Here, \( f(t)=\frac{1}{2+\sqrt{t}} \) and \[ f(t+h)=\frac{1}{2+\sqrt{t+h}}. \] - Thus, \[ f(t+h)-f(t)=\frac{1}{2+\sqrt{t+h}}-\frac{1}{2+\sqrt{t}}. \] - Write this as a single fraction: \[ \frac{(2+\sqrt{t})-(2+\sqrt{t+h})}{(2+\sqrt{t+h})(2+\sqrt{t})}=\frac{\sqrt{t}-\sqrt{t+h}}{(2+\sqrt{t+h})(2+\sqrt{t})}. \] - The difference quotient is then \[ \frac{f(t+h)-f(t)}{h}=\frac{\sqrt{t}-\sqrt{t+h}}{h\,(2+\sqrt{t+h})(2+\sqrt{t})}. \] - Multiply numerator and denominator by the conjugate \(\sqrt{t}+\sqrt{t+h}\) to simplify: \[ (\sqrt{t}-\sqrt{t+h})(\sqrt{t}+\sqrt{t+h})=t-(t+h)=-h. \] - So we have \[ \frac{-h}{h\,(2+\sqrt{t+h})(2+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}. \] - Cancel \( h \): \[ \frac{-1}{(2+\sqrt{t+h})(2+\sqrt{t})(\sqrt{t}+\sqrt{t+h})}. \] - Now, taking the limit \( h\to 0 \) (note that \(\sqrt{t+h}\to\sqrt{t}\)): \[ f'(t)=\frac{-1}{(2+\sqrt{t})^2\,(2\sqrt{t})}. \] - Therefore, \[ \frac{d(f)}{dt}=-\frac{1}{2\sqrt{t}(2+\sqrt{t})^2}. \]