Question

differentiate y = [sin x cos (x²)]/ [ x³ + log x]
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Answer 1

Answer:

*Solution:*

We can differentiate this function using quotient rule, logarithmic-function. When we apply the quotient rule we have to use the product rule in differentiating the numerator. But in the method of logarithmic-differentiation first we have to apply the formulas log(m/n) = log m - log n and log (m n) = log m + log n.

Let y = [sin x cos (x²)]/[ x³ + log x ]

Take log on both sides

log y = log [sin x cos (x²)]/[ x³ + log x ]

log y = log [sin x cos (x²)] - log [ x³ + log x ]

log y = log [sin x ] + log [ cos (x²) ] - log [ x³ + log x ]

(1/y)dy/dx=(1/sin x)cos x-[1/cos (x²)]Sin (x²)(2x)-[1/(x³ + log x )]( 3x² + 1/x )

(1/y)dy/dx=(cos x/sin x)-(2x)sin (x²)/cos (x²)-[1/(x³+log x )]( 3x³ + 1)/x

dy/dx = [cot x - (2x) tan (x²) - ( 3x³ + 1)/x(x³ + log x )]y

dy/dx = [cot x - (2x) tan (x²) - ( 3x³ + 1)/x(x³ + log x )] x

[sin x cos (x²)]/[ x³ + log x ]