Asked by Pat
                Differentiate y=log(8x^2-3x+9)
            
            
        Answers
                    Answered by
            Jai
            
    y = log(8x^2-3x+9)
Note that derivative of log(x) = 1/(x*ln(10))
Thus,
dy/dx
= ( 1/(8x^2-3x+9)*ln(10) ) * (16x - 3)
= (16x - 3) / ((8x^2 - 3x + 9)*ln(10))
Hope this helps~ :3
    
Note that derivative of log(x) = 1/(x*ln(10))
Thus,
dy/dx
= ( 1/(8x^2-3x+9)*ln(10) ) * (16x - 3)
= (16x - 3) / ((8x^2 - 3x + 9)*ln(10))
Hope this helps~ :3
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