What is the derivative of x^x-2^sinx ?

The derivative of x^x-2^sinx is x^x(ln(x)+1)-2^sin

What is the first derivative of x^x-2^sinx ?

The first derivative of x^x-2^sinx is x^x(ln(x)+1)-2^sin

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derivative of

x^{x} - 2^{s} in x

x^{x} (ln (x) + 1) - 2^{s} in

Solution steps

\frac{d}{d x} (x^{x} - 2^{s} in x)

Treat

i, n, s

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d x} (x^{x}) - \frac{d}{d x} (2^{s} in x)

\frac{d}{d x} (x^{x}) = x^{x} (ln (x) + 1)

\frac{d}{d x} (2^{s} in x) = 2^{s} in

= x^{x} (ln (x) + 1) - 2^{s} in

- 0.25 x

\frac{x ^{3} + 1}{x ^{3}}

lo g_{10} (x^{2} + 3)

9.1 1^{x}

2^{- 3 t}

What is the derivative of x^x-2^sinx ?
The derivative of x^x-2^sinx is x^x(ln(x)+1)-2^sin
What is the first derivative of x^x-2^sinx ?
The first derivative of x^x-2^sinx is x^x(ln(x)+1)-2^sin