If y(x)=\sin u, then y' is Option 1:-\frac{du}{dx}\cos u Opt

kuCAu 2021-10-04 Answered
If \(\displaystyle{y}{\left({x}\right)}={\sin{{u}}}\), then y' is
Option 1:\(\displaystyle-{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}{\cos{{u}}}\)
Option 2:\(\displaystyle{\frac{{{d}{u}}}{{{\left.{d}{x}\right.}}}}{\cos{{u}}}\)
Option 3:\(\displaystyle{\cos{{u}}}\)
Option 4:\(\displaystyle-{\cos{{u}}}\)

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Expert Answer

SchepperJ
Answered 2021-10-05 Author has 17330 answers
Step 1: To Find
Given, \(\displaystyle{y}{\left({x}\right)}={\sin{{u}}}\), then we have to find y'.
Step 2: Calculation
Since we have \(\displaystyle{y}{\left({x}\right)}={\sin{{u}}}\), then
\(\displaystyle{y}{\left({u}\right)}={\sin{{u}}}\)
diff. w.r. to u, we get
\(\displaystyle{y}'{\left({u}\right)}={\frac{{{d}}}{{{d}{u}}}}{\left({\sin{{u}}}\right)}\)
\(\displaystyle{y}'{\left({u}\right)}={\cos{{u}}}\)
Hence option (3) is correct.
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