What is the derivative of 2x^{4+7x+5} ?

The derivative of 2x^{4+7x+5} is 2x^{7x+9}(7ln(x)+(7x+9)/x)

What is the first derivative of 2x^{4+7x+5} ?

The first derivative of 2x^{4+7x+5} is 2x^{7x+9}(7ln(x)+(7x+9)/x)

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\frac{d}{d x} (2 x^{4 + 7 x + 5})

2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})

Solution steps

\frac{d}{d x} (2 x^{4 + 7 x + 5})

Simplify

2 x^{4 + 7 x + 5} : 2 x^{7 x + 9}

= \frac{d}{d x} (2 x^{7 x + 9})

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 2 \frac{d}{d x} (x^{7 x + 9})

Apply exponent rule:

a^{b} = e^{b l n (a)}

x^{7 x + 9} = e^{(7 x + 9) l n (x)}

= 2 \frac{d}{d x} (e^{(7 x + 9) l n (x)})

Apply the chain rule:

e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

= e^{(7 x + 9) l n (x)} \frac{d}{d x} ((7 x + 9) ln (x))

\frac{d}{d x} ((7 x + 9) ln (x)) = 7 ln (x) + \frac{7 x + 9}{x}

= 2 e^{(7 x + 9) l n (x)} (7 ln (x) + \frac{7 x + 9}{x})

e^{(7 x + 9) l n (x)} = x^{7 x + 9}

= 2 x^{7 x + 9} (7 ln (x) + \frac{7 x + 9}{x})

\int \frac{y}{( 1 + y ^{2} )} d y

\int \frac{4 x ^{2} - 3 x + 4}{x ^{3} + 4 x} d x

y^{^{''}} + 2 y^{^{'}} + y = 2 e^{- x}

\frac{d}{d x} (36 x^{2} - 72 x)

x \to 0 lim (cos (\frac{π}{x}))

What is the derivative of 2x^{4+7x+5} ?
The derivative of 2x^{4+7x+5} is 2x^{7x+9}(7ln(x)+(7x+9)/x)
What is the first derivative of 2x^{4+7x+5} ?
The first derivative of 2x^{4+7x+5} is 2x^{7x+9}(7ln(x)+(7x+9)/x)