Question

Please help with calculus

Answer 1

For each of these questions, you need to find the derivative $y'$ or $\dfrac{\mathrm dy}{\mathrm dx}$. 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 $m\neq0$, then the slope of any line normal/perpendicular to this line is $-\dfrac1m$.

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$y=\dfrac{6x+3}{3x^2+6x+4}$

The derivative is

$y'=\dfrac{(3x^2+6x+4)(6x+3)'-(6x+3)(3x^2+6x+4)'}{(3x^2+6x+4)^2}=\dfrac{6(3x^2+6x+4)-(6x+3)(6x+6)}{(3x^2+6x+4)^2}$

$y'=\dfrac{6-18x-18x^2}{(3x^2+6x+4)^2}$

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

$y'=\dfrac{6-18-18}{(3+6+4)^2}=-\dfrac{30}{169}$

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

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 $-\dfrac1{11}$. The point-slope and slope-intercept forms of this line are

$y-9=-\dfrac1{11}(x-1)\implies y=-\dfrac x{11}+\dfrac{100}{11}$

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$y=9x-14$

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

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