What is the derivative of x+axIn(bx) ?

The derivative of x+axIn(bx) is 1+2abInx

What is the first derivative of x+axIn(bx) ?

The first derivative of x+axIn(bx) is 1+2abInx

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\frac{d}{d x} (x + a x I n (b x))

1 + 2 ab I n x

Solution steps

\frac{d}{d x} (x + a x I n (b x))

Simplify

x + a x I n (b x) : x + abn x^{2} I

= \frac{d}{d x} (x + abn x^{2} I)

Treat

a, n, b, I

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d x}{d x} + \frac{d}{d x} (abn x^{2} I)

\frac{d x}{d x} = 1

\frac{d}{d x} (abn x^{2} I) = 2 ab I n x

= 1 + 2 ab I n x

y^{^{''}} + 25 y = 5 t^{2}

\frac{8}{7 x ^{9} + 3}

x \to 0 + lim (x^{7 x})

\frac{d}{d x} (\frac{2 x - 3}{4 x + 1})

\int \frac{( x ^{2} )}{2 ^{x}} d x

What is the derivative of x+axIn(bx) ?
The derivative of x+axIn(bx) is 1+2abInx
What is the first derivative of x+axIn(bx) ?
The first derivative of x+axIn(bx) is 1+2abInx