What is the derivative of 0.7x^35^x ?

The derivative of 0.7x^35^x is 0.7(3*5^xx^2+1.60943…*5^xx^3)

What is the first derivative of 0.7x^35^x ?

The first derivative of 0.7x^35^x is 0.7(3*5^xx^2+1.60943…*5^xx^3)

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derivative of

0.7 x^{3} 5^{x}

0.7 (3 \cdot 5^{x} x^{2} + 1.60943 \dots \cdot 5^{x} x^{3})

Solution steps

\frac{d}{d x} (0.7 x^{3} \cdot 5^{x})

Take the constant out:

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

= 0.7 \frac{d}{d x} (x^{3} \cdot 5^{x})

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{3}, g = 5^{x}

= 0.7 (\frac{d}{d x} (x^{3}) 5^{x} + \frac{d}{d x} (5^{x}) x^{3})

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

\frac{d}{d x} (5^{x}) = 5^{x} ln (5)

= 0.7 (3 x^{2} \cdot 5^{x} + 5^{x} ln (5) x^{3})

Refine to a decimal form

= 0.7 (3 \cdot 5^{x} x^{2} + 1.60943 \dots \cdot 5^{x} x^{3})

\int sinh (d) x d x

t an g e n t f (x) = 5 x + 6, at x = 6

5 \cdot \frac{d Q}{d t} + \frac{Q}{3} = 0

f (x) = 5 e^{x}

x^{2} \frac{d y}{d x} + x y = x^{3}

What is the derivative of 0.7x^35^x ?
The derivative of 0.7x^35^x is 0.7(3*5^xx^2+1.60943…*5^xx^3)
What is the first derivative of 0.7x^35^x ?
The first derivative of 0.7x^35^x is 0.7(3*5^xx^2+1.60943…*5^xx^3)