Derivative of (2+3sinx)(4+5cosx)tanx

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SUMMARY

The derivative of the function f(x) = (2 + 3sin(x))(4 + 5cos(x))tan(x) is calculated using the product rule. The correct derivative is f'(x) = (4 + 5)tan(x)(3cos(x)) + (2 + 3sin(x))(tan(x)(-5sin(x)) + sec²(x)(4 + 5cos(x))). This solution was confirmed by forum participants, with one user noting that WolframAlpha applies the product rule differently, leading to a potential simplification discrepancy.

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Homework Statement



Derive the following:
f(x)= (2+3sinx)(4+5cosx)tanx



2. The attempt at a solution
I derive it and got
f'(x)=(4+5)tanx(3cosx)+(2+3sinx)(tanx(-5sinx)+sec2x(4+5cosx))

Can someone confirm if this is right?
 
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Correct, good work
 
Thanks!
I put it first on WolframAlpha to compare, but it does the product rule the other way I did so I didnt felt like simplifying.
 

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