Question

Please help with calculus

Answer 1

For each of these questions, you need to find the derivative
`y'` or
`(\mathrm dy)/(\mathrm dx)`. The slope of the tangent to these curves at the point
`(a,b)` is the value of
`y'` when
`x=a` and
`y=b`. It's also important to know that if the slope of a line is
`m\\eq0`, then the slope of any line normal/perpendicular to this line is
`-\frac1m`.

###

`y=(6x+3)/(3x^2+6x+4)`

The derivative is

`y'=((3x^2+6x+4)(6x+3)'-(6x+3)(3x^2+6x+4)')/((3x^2+6x+4)^2)=(6(3x^2+6x+4)-(6x+3)(6x+6))/((3x^2+6x+4)^2)`

`y'=(6-18x-18x^2)/((3x^2+6x+4)^2)`

When
`x=1`, we get a slope of

`y'=(6-18-18)/((3+6+4)^2)=-(30)/(169)`

###

`y=x^2+9x+16`

The derivative is

`y'=2x+9`

and so the tangent line at (1, 9) has slope

`y'=2+9=11`

The line normal to this has slope
`-\frac1{11}`. The point-slope and slope-intercept forms of this line are

`y-9=-\frac1{11}(x-1)\implies y=-\frac x{11}+(100)/(11)`

###

`y=9x-14`

The derivative is

`y'=9`

so the slope of any line tangent to the curve is 9. The line that passes through (3, 4) is

`y-4=9(x-3)\impleis y=9x-23`