Question

a^2 + b^2 - c^2 + 2ab​

Answer 1

Answer:

ok but mark

Step-by-step explanation:

FORMULAE OF DIFFERENTIATION ⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

→\tt{\dfrac{d}{dx} x^{n} = nx^{n-1}}

dx

d

x

n

=nx

n−1

→\tt{\dfrac{d}{dx} (constant) = 0} ⠀

dx

d

(constant)=0⠀

→ \tt{\dfrac{d}{dx} kf(x) = k. \dfrac{d}{dx} f(x)}

dx

d

kf(x)=k.

dx

d

f(x)

→ \tt{\dfrac{d}{dx} (u+v) = \dfrac{du}{dx} + \dfrac{dv}{dx} }

dx

d

(u+v)=

dx

du

+

dx

dv

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀

→\tt{\dfrac{d}{dx} (u-v) = \dfrac{du}{dx} - \dfrac{dv}{dx}}

dx

d

(u−v)=

dx

du

−

dx

dv

⠀⠀⠀⠀⠀⠀⠀⠀⠀

→ \tt{\dfrac{d}{dx} (u.v) = u \dfrac{dv}{dx} + v \dfrac{du}{dx}}

dx

d

(u.v)=u

dx

dv

+v

dx

du

⠀⠀⠀⠀⠀⠀

→ \tt{\dfrac{d}{dx} (\dfrac{u}{v}) = \dfrac{v \dfrac{du}{dx} - u \dfrac{dv}{dx}}{v^2}}

dx

d

(

v

u

)=

v

2

v

dx

du

−u

dx

dv

→\tt{\dfrac{d}{dx} (Cos x) = - sin x}

dx

d

(Cosx)=−sinx

→\tt{\dfrac{d}{dx} (Sin x) = Cos x}

dx

d

(Sinx)=Cosx

→\tt{\dfrac{d}{dx} (Tan x) = Sec^2 x}

dx

d

(Tanx)=Sec

2

x

→\tt{\dfrac{d}{dx} (Cot x) = - Cosec^2 x}

dx

d

(Cotx)=−Cosec

2

x

→\tt{\dfrac{d}{dx} (Sec x) = Sec x. Tan x}

dx

d

(Secx)=Secx.Tanx

→\tt{\dfrac{d}{dx} (Cosec x) = - Cosec x. Cot x}

dx

d

(Cosecx)=−Cosecx.Cotx ⠀

→\tt{\dfrac{d}{dx} log_{e}(x) = \dfrac{1}{x}}

dx

d

log

e

(x)=

x

1

→ \tt{\dfrac{d}{dx} e^x = e^x}

dx

d

e

x

=e

x

⠀⠀⠀⠀⠀

→\tt{\dfrac{d}{dx} a^x = a^{x} . log_{e}{a}}

dx

d

a

x

=a

x

.log

e