Find-the-derivatives-f-x-of-the-following-function-with-respect-to-x-f-x-Sin-pi-Sinx-pi-Cosx-Mastermind-

Question Number 180839 by Mastermind last updated on 17/Nov/22

Find the derivatives f^{^{'}} (x) of the

following function with respect to x :

f (x) = Sin (π^{Sinx} + π^{Cosx}) .

Mastermind

Answered by saboorhalimi last updated on 17/Nov/22

Solution by : Saboor Halimi

f (x) = Sin (π^{sinx} + π^{\cos (x)}

f^{'} (x) = (π^{\sin (x)} \cos (x) \ln (π) - π^{cosx} sinxln (π)) \cos (π^{\sin (x)} + π^{\cos (x)})

Commented by Mastermind last updated on 17/Nov/22

With full details explanation pls

Commented by Acem last updated on 17/Nov/22

{(e^{x})}^{'} = x^{'} e^{x} \ln e = 1 \times e^{x} \times 1 = e^{x}

{(a^{u})}^{'} = \ln a \times u^{'} \times a^{u} h e r e a = π

{(\cos u)}^{'} = - u^{'} \sin u

{(\sin u)}^{'} = u^{^{'}} \cos u

{[\sin (π^{\sin x} + π^{\cos x})]}^{^{'}} = {(π^{\sin x} + π^{\cos x})}^{'} . \cos (π^{\sin x} + π^{\cos x})

{(π^{\sin x} + π^{\cos x})}^{'} = {(π^{\sin x})}^{^{'}} + {(π^{\cos x})}^{'} {(a^{u})}^{'} l o o k a b o v e

= \ln π . \cos x π^{\sin x} - \ln π . \sin x . π^{\cos x}

f {(x)}^{'} = \ln π (\cos x π^{\sin x} - \sin x . π^{\cos x}) \cos (π^{\sin x} + π^{\cos x})

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