Question
\(\frac{d}{dx}(sin^{-1}(x^2))|_{x=\frac{1}{4}}\)
A \(\frac{2(\frac{1}{4})}{1+(\frac{1}{4})^4}\)
B \(\frac{2(\frac{1}{4})}{\sqrt{1−(\frac{1}{4})^4}}\)
C \(2(\frac{1}{4})cos^{−1}(\frac{1}{16})\)
D \(−2(\frac{1}{4})cot(\frac{1}{16})csc(\frac{1}{16})\)
Answer/Explanation
Ans:B
Question
\(\frac{d}{dx}(cos^{−1}(−3x))=\)
A \(\frac{3}{\sqrt{1−(−3x)^2}}\)
B \(\frac{−3}{\sqrt{1−(−3x)^2}}\)
C \(−sin^{−1}(−3x)⋅(−3)\)
D \(−cos^{−2}(−3x)⋅(−3)\)
Answer/Explanation
Ans:A
Question
Which of the following methods can be used to find the derivative of \(y=arccos(\sqrt{x})\) with respect to x ?
I. Use the quotient rule to differentiate \(\frac{1}{cos(\sqrt{x})}\) .
II. Use the chain rule to differentiate \(cos(arccos(\sqrt{x}))=\sqrt{x}\) .
III. Use implicit differentiation to differentiate the function y in the relation \(cosy=\sqrt{x}\) with respect to x.
A. I only
B. III only
C. II and III only
D. I, II, and III
Answer/Explanation
Ans: C
The definition of the inverse cosine function means that \(cos(arccos(\sqrt{x}))=\sqrt{x}\) . Using method II, \(\frac{d}{dx}cos(arccos(\sqrt{x}))=−sin(arccos(\sqrt{x}))⋅(\frac{d}{dx}arccos(\sqrt{x}))=\frac{d}{dx}(\sqrt{x})=\frac{1}{2\sqrt{x}}⇒\frac{d}{dx}(arccos(\sqrt{x}))=\frac{−1}{2\sqrt{x}sin(arccos(\sqrt{x}))}\). If \(θ=arccos(\sqrt{x})\), then \(0≤θ≤\frac{π}{2}\) and \(cosθ=\sqrt{x}\). Therefore, \(sinθ\) is non-negative and \(sin^2θ=1−cos^2θ⇒sinθ=\sqrt{1−cos^2θ}=\sqrt{1−(\sqrt{x})^2}=\sqrt{1−x}\) because \(0≤x≤1\). The derivative is then \(\frac{d}{dx}arccos(\sqrt{x})=\frac{-1}{2\sqrt{x}.sin(arccos\sqrt{x}))}=\frac{-1}{2\sqrt{x}.sin\theta }=\frac{-1}{2\sqrt{x}\sqrt{1-\sqrt{x}}}=\frac{-1}{2\sqrt{x(1-x)}}\) for \(0<x<1\).
Question
What are the coordinates of the inflection point on the graph of y=(x+1) arctan ?
(A) (−1,0) (B) (0,0 ) (C) (0,1) (D) \(\left ( 1,\frac{\pi }{4} \right )\) (E) \(\left ( 1,\frac{\pi }{2} \right )\)