What is the derivative of y=(7)(6^{sqrt(x-8)}) ?

The derivative of y=(7)(6^{sqrt(x-8)}) is (7ln(6)*3^{sqrt(x-8)}*2^{sqrt(x-8)-1})/(sqrt(x-8))

What is the first derivative of y=(7)(6^{sqrt(x-8)}) ?

The first derivative of y=(7)(6^{sqrt(x-8)}) is (7ln(6)*3^{sqrt(x-8)}*2^{sqrt(x-8)-1})/(sqrt(x-8))

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derivative of

y = (7) (6^{x - 8})

\frac{7 ln ( 6 ) \cdot 3 ^{x - 8} \cdot 2 ^{x - 8 - 1}}{x - 8}

Solution steps

\frac{d}{d x} (7 \cdot 6^{x - 8})

Take the constant out:

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

= 7 \frac{d}{d x} (6^{x - 8})

Apply exponent rule:

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

6^{x - 8} = e^{x - 8 l n (6)}

= 7 \frac{d}{d x} (e^{x - 8 l n (6)})

Apply the chain rule:

e^{x - 8 l n (6)} \frac{d}{d x} (x - 8 ln (6))

= e^{x - 8 l n (6)} \frac{d}{d x} (x - 8 ln (6))

\frac{d}{d x} (x - 8 ln (6)) = \frac{ln ( 6 )}{2 x - 8}

= 7 e^{x - 8 l n (6)} \frac{ln ( 6 )}{2 x - 8}

Simplify

7 e^{x - 8 l n (6)} \frac{ln ( 6 )}{2 x - 8} : \frac{7 ln ( 6 ) \cdot 3 ^{x - 8} \cdot 2 ^{x - 8 - 1}}{x - 8}

= \frac{7 ln ( 6 ) \cdot 3 ^{x - 8} \cdot 2 ^{x - 8 - 1}}{x - 8}

x^{2} \frac{d y}{d x} + 4 x y - e^{x} = 0

(2 y - 8 x^{2}) d x + x d y = 0

\frac{d}{d x} (e^{x \cdot s i n (x)})

x \to 2 + lim (\frac{1}{x - 2})

f (x) = (arcsin (5 x + 3))^{3}

What is the derivative of y=(7)(6^{sqrt(x-8)}) ?
The derivative of y=(7)(6^{sqrt(x-8)}) is (7ln(6)*3^{sqrt(x-8)}*2^{sqrt(x-8)-1})/(sqrt(x-8))
What is the first derivative of y=(7)(6^{sqrt(x-8)}) ?
The first derivative of y=(7)(6^{sqrt(x-8)}) is (7ln(6)*3^{sqrt(x-8)}*2^{sqrt(x-8)-1})/(sqrt(x-8))