I got $\frac{\sec^2(\arcsin t)}{\sqrt{1-t^2}}$, however my textbook shows $\frac{1}{(1-t^2)^{3/2}}$. Would my answer be acceptable on a test?
If not, how do I get to the correct answer?
Thanks
Is this answer acceptable for the derivative of $\tan(\arcsin t)$?
0
$\begingroup$
calculus
derivatives
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2Well $csc^2(x)=\frac{1}{1-\sin^2x}$ – 2017-01-07
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1I think you ought to simplify first. Note that $\tan=\frac{\sin}{\cos}$ and $\cos(\arcsin t)=\sqrt{1-t^2}$. – 2017-01-07
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0Every teacher has his own ways. I would take points off if you didn't simplify that. I let some simplifications go, but this one matters. I consider it part of the required material that calculus student can simplify trig(arctrig) things. – 2017-01-07
2 Answers
1
Your answer is correct, but notice that: $$ \sec^2(\arcsin t) = \frac{1}{\cos^2(\arcsin t)} = \frac{1}{1-\sin^2(\arcsin t)} = \frac{1}{1-t^2} $$
3
Remember the following identities:
$$\sec^2(z)=\frac 1{\cos^2 z}$$
$$\cos \arcsin z = {\sqrt {1-z^2}}$$
substitution in your case gave the desired equality.