What is (\partial)/(\partial x)((ln^'(x^2+5))/(y^2)) ?

The answer to (\partial)/(\partial x)((ln^'(x^2+5))/(y^2)) is (2'xln^{'-1}(x^2+5))/(y^2(x^2+5))

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(\partial)/(\partial x)((ln^'(x^2+5))/(y^2))

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

\frac{2' x ln ^{' - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}

Solution steps

\frac{\partial}{\partial x} (\frac{ln ^{^{'}} ( x ^{2} + 5 )}{y ^{2}})

Treat

y

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{1}{y ^{2}} \frac{\partial}{\partial x} (ln^{^{^{'}}} (x^{2} + 5))

Apply exponent rule:

a^{b} = e^{b l n (a)}

ln^{^{'}} (x^{2} + 5) = e^{^{'} l n (l n (x^{2} + 5))}

= \frac{1}{y ^{2}} \frac{\partial}{\partial x} (e^{^{^{'}} l n (l n (x^{2} + 5))})

Apply the chain rule:

e^{^{'} l n (l n (x^{2} + 5))} \frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5)))

= e^{^{'} l n (l n (x^{2} + 5))} \frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5)))

\frac{\partial}{\partial x} (^{'} ln (ln (x^{2} + 5))) = \frac{2 ^{'} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )}

= \frac{1}{y ^{2}} e^{^{^{'}} l n (l n (x^{2} + 5))} \frac{2 ^{^{'}} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )}

Simplify

\frac{1}{y ^{2}} e^{^{'} l n (l n (x^{2} + 5))} \frac{2 ^{'} x}{ln ( x ^{2} + 5 ) ( x ^{2} + 5 )} : \frac{2 ^{'} x ln ^{^{'} - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}

= \frac{2 ^{'} x ln ^{^{'} - 1} ( x ^{2} + 5 )}{y ^{2} ( x ^{2} + 5 )}

\frac{\partial}{\partial z} (x y arcsin (yz))

\frac{d}{d x} (3 + \frac{1}{2} cosh (2 x))

f (x) = x^{2} - 9 x + 17, at x = 4

x^{3} - 3 x + 3

\frac{8 x}{x ^{2} + 3}

What is (\partial)/(\partial x)((ln^'(x^2+5))/(y^2)) ?
The answer to (\partial)/(\partial x)((ln^'(x^2+5))/(y^2)) is (2'xln^{'-1}(x^2+5))/(y^2(x^2+5))