What is the first derivative of $x + tan x$ $x+ tanx$ and how do you find the slope of the normal to the curve where $x = π$ $x=pi$ ?

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Answer 1 · 2015-08-21T08:59:47

The slope of the normal line at that point is $- \frac{1}{2}$ $-1/2$

$\frac{d y}{d x} = 1 + \frac{1}{{cos}^{2} x}$ $dy/dx=1+1/cos^2x$

Therefore:

${(\frac{d y}{d x})}_{x = π} = 1 + \frac{1}{{cos}^{2} π} = 1 + {(- 1)}^{2} = 2$ $(dy/dx)_(x=pi)=1+1/cos^2pi=1+(-1)^2=2$

So the slope of the tangent line is $2$ $2$ and the slope of the normal line is $- \frac{1}{2}$ $-1/2$

What is the first derivative of x+tanxx+ tanx and how do you find the slope of the normal to the curve where x=πx=pi?