Question

find the slope of the line tangent to ( y=arctan (2 x) ) at ( x = 1 ).
hint: ( arctan (2 x)=\tan ^{-1}(2 x) )
( \frac{1}{2} )
( \frac{1}{5} )
( \frac{2}{5} )
4
answer: type your answer...
2 fill in the blank 1 point
find the slope of the curve ( x y^{2}+x = 1 ) at ( left(\frac{1}{2}, 2
ight) ).
-5
-2
( -\frac{5}{2} )
( -\frac{1}{4} )
0
answer: type your answer...

Explanation:

Step1: Differentiate \(y = \arctan(2x)\)

Use the chain rule. The derivative of \(\arctan(u)\) is \(\frac{1}{1 + u^{2}}\cdot u'\). Here \(u = 2x\), so \(u'=2\). Then \(y'=\frac{2}{1+(2x)^{2}}=\frac{2}{1 + 4x^{2}}\)

Step2: Evaluate the derivative at \(x = 1\)

Substitute \(x = 1\) into \(y'\). \(y'(1)=\frac{2}{1+4\times1^{2}}=\frac{2}{5}\)

Step1: Differentiate \(xy^{2}+x = 1\) implicitly with respect to \(x\)

Using the product rule \((uv)'=u'v + uv'\) (where \(u = x\) and \(v = y^{2}\)), we get \(y^{2}+2xyy'+1 = 0\)

Step2: Solve for \(y'\)

\(2xyy'=-y^{2}-1\), so \(y'=\frac{-y^{2}-1}{2xy}\)

Step3: Substitute \(x=\frac{1}{2}\) and \(y = 2\) into \(y'\)

\(y'=\frac{-2^{2}-1}{2\times\frac{1}{2}\times2}=\frac{-4 - 1}{2}=-\frac{5}{2}\)

Answer:

c. \(\frac{2}{5}\)