Loading Student:-Calculus1

$f\left(t\right)\=\frac{t-1}{t^2+2}$$\stackrel{\text{assign as function}}{\to }$$f$

$f\prime \left(t\right)$ = $\frac{1}{t^2+2}-\frac{2\left(t-1\right)t}{{\left(t^2+2\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{-t^2+2t+2}{{\left(t^2+2\right)}^2}$

Annotated stepwise solution

$\frac{\ⅆ}{\ⅆ t}\left(f\left(t\right)\right)$$\stackrel{\text{show solution steps}}{\to }$$\begin{array}{lll}& & \text{Differentiation Steps}\\ & & \frac{\ⅆ}{\ⅆt}\left(\frac{t-1}{t^2+2}\right)\\ \text{▫}& & \text{1. Apply the}\mathbf{\text{quotient}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{quotient}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}\left(\frac{t-1}{\left(t^2+2\right)g\left(t\right)}\right)=\frac{\frac{\ⅆ}{\ⅆt}\left(\frac{t-1}{t^2+2}\right)g\left(t\right)-\frac{\left(x-1\right)\frac{\ⅆ}{\ⅆt}g\left(t\right)}{x^2+2}}{{g\left(t\right)}^2}\\ & & f\left(t\right)=t-1\\ & & g\left(t\right)=t^2+2\\ & & \text{This gives:}\\ & & \frac{\frac{\ⅆ}{\ⅆt}\left(t-1\right)\left(t^2+2\right)-\left(t-1\right)\frac{\ⅆ}{\ⅆt}\left(t^2+2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{2. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}\left(f\left(t\right)+g\left(t\right)\right)=\frac{\ⅆ}{\ⅆt}f\left(t\right)+\frac{\ⅆ}{\ⅆt}g\left(t\right)\\ & & f\left(t\right)=t\\ & & g\left(t\right)=\mathrm{-1}\\ & & \text{This gives:}\\ & & \frac{\left(\frac{\ⅆ}{\ⅆt}t+\frac{\ⅆ}{\ⅆt}\left(\mathrm{-1}\right)\right)\left(t^2+2\right)-\left(t-1\right)\frac{\ⅆ}{\ⅆt}\left(t^2+2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{3. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆt}\left(\mathrm{-1}\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}C=0\\ & \text{◦}& \text{This means}\\ & & \frac{\ⅆ}{\ⅆt}\left(\mathrm{-1}\right)=0\\ & & \text{We can now rewrite the derivative as:}\\ & & \frac{\frac{\ⅆ}{\ⅆt}t\left(t^2+2\right)-\left(t-1\right)\frac{\ⅆ}{\ⅆt}\left(t^2+2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{4. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆt}t\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}\left(t^n\right)=n{t}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆt}{t}^1=1\cdot t^0\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆt}t=1\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{t^2+2-\left(t-1\right)\frac{\ⅆ}{\ⅆt}\left(t^2+2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{5. Apply the}\mathbf{\text{sum}}\text{rule}\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{sum}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}\left(f\left(t\right)+g\left(t\right)\right)=\frac{\ⅆ}{\ⅆt}f\left(t\right)+\frac{\ⅆ}{\ⅆt}g\left(t\right)\\ & & f\left(t\right)=t^2\\ & & g\left(t\right)=2\\ & & \text{This gives:}\\ & & \frac{t^2+2-\left(t-1\right)\left(\frac{\ⅆ}{\ⅆt}\left(t^2\right)+\frac{\ⅆ}{\ⅆt}2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{6. Apply the}\mathbf{\text{constant}}\text{rule to the term}\frac{\ⅆ}{\ⅆt}2\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{constant}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}C=0\\ & \text{◦}& \text{This means}\\ & & \frac{\ⅆ}{\ⅆt}2=0\\ & & \text{We can now rewrite the derivative as:}\\ & & \frac{t^2+2-\left(t-1\right)\frac{\ⅆ}{\ⅆt}\left(t^2\right)}{{\left(t^2+2\right)}^2}\\ \text{▫}& & \text{7. Apply the}\mathbf{\text{power}}\text{rule to the term}\frac{\ⅆ}{\ⅆt}\left(t^2\right)\\ & \text{◦}& \text{Recall the definition of the}\mathbf{\text{power}}\text{rule}\\ & & \frac{\ⅆ}{\ⅆt}\left(t^n\right)=n{t}^{n-1}\\ & \text{◦}& \text{This means:}\\ & & \frac{\ⅆ}{\ⅆt}\left(t^2\right)=2\cdot t^1\\ & \text{◦}& \text{So,}\\ & & \frac{\ⅆ}{\ⅆt}\left(t^2\right)=2\cdot t\\ & & \text{We can rewrite the derivative as:}\\ & & \frac{t^2+2-2\left(t-1\right)t}{{\left(t^2+2\right)}^2}\end{array}$

Interactive solution by Differentiation Methods tutor