What is (\partial)/(\partial x)(({f}(x))^') ?

The answer to (\partial)/(\partial x)(({f}(x))^') is '(\partial)/(\partial x)({f}(x)){f}(x)^{'-1}

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(\partial)/(\partial x)(({f}(x))^')

\frac{\partial}{\partial x} ((f (x))^{^{'}})

' \frac{\partial}{\partial x} (f (x)) f (x)^{' - 1}

Solution steps

\frac{\partial}{\partial x} ((f (x))^{^{'}})

Apply exponent rule:

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

f (x)^{^{'}} = e^{^{'} l n (f (x))}

= \frac{\partial}{\partial x} (e^{^{^{'}} l n (f (x))})

Apply the chain rule:

e^{^{'} l n (f (x))} \frac{\partial}{\partial x} (^{'} ln (f (x)))

= e^{^{'} l n (f (x))} \frac{\partial}{\partial x} (^{'} ln (f (x)))

\frac{\partial}{\partial x} (^{'} ln (f (x))) = \frac{^{'} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )}

= e^{^{^{'}} l n (f (x))} \frac{^{^{'}} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )}

Simplify

e^{^{'} l n (f (x))} \frac{^{'} \frac{\partial}{\partial x} ( f ( x ) )}{f ( x )} :^{'} \frac{\partial}{\partial x} (f (x)) f (x)^{^{'} - 1}

=^{'} \frac{\partial}{\partial x} (f (x)) f (x)^{^{'} - 1}

x y^{^{''}} + y^{^{'}} = x

\frac{3 x + 4}{7 x - 5}

a y^{^{'}} + b y = 0

4 t x \frac{d x}{d t} = x^{2} + 1

\frac{\partial}{\partial z} (cos (k z - wt))

What is (\partial)/(\partial x)(({f}(x))^') ?
The answer to (\partial)/(\partial x)(({f}(x))^') is '(\partial)/(\partial x)({f}(x)){f}(x)^{'-1}