find the slope of the line tangent to ( y=arctan (2 x) ) at ( x = 1 ).\nhint: ( arctan (2 x)=\tan ^{-1}(2 x)…

find the slope of the line tangent to ( y=arctan (2 x) ) at ( x = 1 ).\nhint: ( arctan (2 x)=\tan ^{-1}(2 x) )\n( \frac{1}{2} )\n( \frac{1}{5} )\n( \frac{2}{5} )\n4\nanswer: type your answer...\n2 fill in the blank 1 point\nfind the slope of the curve ( x y^{2}+x = 1 ) at ( left(\frac{1}{2}, 2\right) ).\n-5\n-2\n( -\frac{5}{2} )\n( -\frac{1}{4} )\n0\nanswer: type your answer...

find the slope of the line tangent to ( y=arctan (2 x) ) at ( x = 1 ).\nhint: ( arctan (2 x)=\tan ^{-1}(2 x) )\n( \frac{1}{2} )\n( \frac{1}{5} )\n( \frac{2}{5} )\n4\nanswer: type your answer...\n2 fill in the blank 1 point\nfind the slope of the curve ( x y^{2}+x = 1 ) at ( left(\frac{1}{2}, 2\right) ).\n-5\n-2\n( -\frac{5}{2} )\n( -\frac{1}{4} )\n0\nanswer: type your answer...

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})

Answer:

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

Explanation:

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{5}{2})