What is the derivative of f(x)=(3x^5+x6^x)2^{32} ?

The derivative of f(x)=(3x^5+x6^x)2^{32} is 2^{32}(15x^4+6^x+ln(6)*6^xx)

What is the first derivative of f(x)=(3x^5+x6^x)2^{32} ?

The first derivative of f(x)=(3x^5+x6^x)2^{32} is 2^{32}(15x^4+6^x+ln(6)*6^xx)

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derivative of

f (x) = (3 x^{5} + x 6^{x}) 2^{32}

2^{32} (15 x^{4} + 6^{x} + ln (6) \cdot 6^{x} x)

Solution steps

\frac{d}{d x} ((3 x^{5} + x \cdot 6^{x}) \cdot 2^{32})

Take the constant out:

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

= 2^{32} \frac{d}{d x} (3 x^{5} + x \cdot 6^{x})

Apply the Sum/Difference Rule:

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

= 2^{32} (\frac{d}{d x} (3 x^{5}) + \frac{d}{d x} (x \cdot 6^{x}))

\frac{d}{d x} (3 x^{5}) = 15 x^{4}

\frac{d}{d x} (x \cdot 6^{x}) = 6^{x} + ln (6) \cdot 6^{x} x

= 2^{32} (15 x^{4} + 6^{x} + ln (6) \cdot 6^{x} x)

(1, - 9), (- 3, - 9)

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What is the derivative of f(x)=(3x^5+x6^x)2^{32} ?
The derivative of f(x)=(3x^5+x6^x)2^{32} is 2^{32}(15x^4+6^x+ln(6)*6^xx)
What is the first derivative of f(x)=(3x^5+x6^x)2^{32} ?
The first derivative of f(x)=(3x^5+x6^x)2^{32} is 2^{32}(15x^4+6^x+ln(6)*6^xx)