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Calculus/normal and tangent lines

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Expert: - 3/18/2010


Question
What is the slope of the line tangent to the curve y = arctan (4x) when x = ¼ ?

Answer
Ii can be done it by 4x = tan(y).
Taking the derivative gives 4 dx = sec²y dy.

Thig can be solved for dy/dx just like any other variable.
Divide both sides by dx and both sides by sec²y, giving 4/sec²y = dy/dx.

Now since tan(4x) = 4x/1, the far side of a right triangle can be made as 4x and
the near side as 1.  Since (4x)² = 16x², this makes the hypoteneuse into √(1+16x²).
I used to be unclear about how to do the sec(), but I remembered it was 1/cos().
Since the cos(y) = 1/√(1+16x²), the sec(y)= √(1+16x²).

This can be put back in the formula for dy/dx.
Note that sec() is squared in the equation, the squareroot sign goes away.

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