What is ln(((sin(x^2))')/((cos(2x))')) ?

The solution to ln(((sin(x^2))')/((cos(2x))')) is ln((sin(x^2))')-ln((cos(2x))')

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simplify ln(((sin(x^2))')/((cos(2x))'))

simplify

ln (\frac{( sin ( x ^{2} ) ) ^{'}}{( cos ( 2 x ) ) ^{'}})

ln ((sin (x^{2}))') - ln ((cos (2 x))')

Solution steps

ln (\frac{( sin ( x ^{2} ) ) ^{'}}{( cos ( 2 x ) ) ^{'}})

Apply log rule:

lo g_{c} (\frac{a}{b}) = lo g_{c} (a) - lo g_{c} (b)

ln (\frac{( sin ( x ^{2} ) ) ^{'}}{( cos ( 2 x ) ) ^{'}}) = ln ((sin (x^{2}))^{'}) - ln ((cos (2 x))^{'})

= ln ((sin (x^{2}))^{^{'}}) - ln ((cos (2 x))^{^{'}})

- ln (10) + ln (5)

ln (\frac{3 x - 2 y}{2 x + 2 y})

- lo g_{10} (1.58 \cdot 1 0^{- 5})

ln (\frac{e ^{x^{2}} \cdot cos ( x )}{2 x + 4})

ln (\frac{\frac{π}{4} \cdot 4 7 ^{2}}{\frac{π}{4} \cdot 3 0 ^{2}})

What is ln(((sin(x^2))')/((cos(2x))')) ?
The solution to ln(((sin(x^2))')/((cos(2x))')) is ln((sin(x^2))')-ln((cos(2x))')