What is the derivative of Ax^n+Be^x ?

The derivative of Ax^n+Be^x is Anx^{n-1}+Be^x

What is the first derivative of Ax^n+Be^x ?

The first derivative of Ax^n+Be^x is Anx^{n-1}+Be^x

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derivative of

A x^{n} + B e^{x}

A n x^{n - 1} + B e^{x}

Solution steps

\frac{d}{d x} (A x^{n} + B e^{x})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (A x^{n}) + \frac{d}{d x} (B e^{x})

\frac{d}{d x} (A x^{n}) = A n x^{n - 1}

\frac{d}{d x} (B e^{x}) = B e^{x}

= A n x^{n - 1} + B e^{x}

\frac{d}{d x} (\frac{x ^{2}}{x - 3})

(6 x + 5) (x^{2} - 3 x)

\int 9 x^{5} d x

\int sin (r^{2}) d r

\int cos (2 π t) d t

What is the derivative of Ax^n+Be^x ?
The derivative of Ax^n+Be^x is Anx^{n-1}+Be^x
What is the first derivative of Ax^n+Be^x ?
The first derivative of Ax^n+Be^x is Anx^{n-1}+Be^x