The derivative of f(x)=\frac{3 \sin x}{2+\cos x} My solution: The background 3 \frac{d}{dx}(\frac{\sin

David Young

David Young

Answered question

2022-01-16

The derivative of f(x)=3sinx2+cosx
My solution:
The background
3ddx(sinx2+cos(x))
=3ddx(sin(x))(2+cos(x))ddx(2+cos(x))sin(x)(2+cos(x))2
=3cos(x)(2+cos(x))(sin(x))sinx(2+cos(x))2
=3+6cos(x)(2+cos(x))2
However, I plugged f(x)=3sinx2+cosx into the derivative calculator in wolfram alpha and received the following calculation:
3(sin2(x)+cos2(x)+2cos(x))(2+cos(x))2
Is my solution incorrect and the one from wolfram alpha correct?

Answer & Explanation

Cassandra Ramirez

Cassandra Ramirez

Beginner2022-01-17Added 30 answers

Your answer is correct. Use trig. identity sin2(x)+cos2(x)=1
3(sin2(x)+cos2(x)+2cos(x))(2+cos(x))2=3(1+2cos(x))(2+cos(x))2=3+6cos(x)(2+cos(x))2
Above matches your solution
Lakisha Archer

Lakisha Archer

Beginner2022-01-18Added 39 answers

They are identical. The numerator of wolfram alpha
3(sin2x+cos2x+2cos(x))
=3(1+2cosx)=3+6cosx
which is the same as the numerator of derivative you correctly worked out.

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