Question

derivative of (a^x/Log a)?​

Answer 1

$\large\bold{\underline{\underline{To \: \: Differentive:-}}}$

$\sf{\dfrac{a^x}{log(a)} }$

(OR)

$\sf{\dfrac{a^x}{ln(a)} }$

$\large\bold{\underline{\underline{Solution:-}}}$

$\sf{Let\ y=\dfrac{a^x}{ln(a)} }$

$\sf{=\dfrac{d}{dx}\bigg[\dfrac{a^x}{ln(a)} \bigg] }$

(Linear differentiation)

$\sf{=\dfrac{1}{ln(a)}.\dfrac{d}{dx}(a^x) }$

$\sf{=\dfrac{ln(a)a^x}{ln(a)} }$

$\boxed{\sf{=a^x}}$

$\large\bold{\underline{\underline{Applied:-}}}$

✭ Linear differentiation

→ $\sf{[a.b(x)+c.d(x)]'=a.b'(x)+c.d'(x)}$

✭ Exponential function rule

→ $\sf{(a^x)'=ln(a).a^x}$

Answer 2

Answer:

a^x.

Step-by-step explanation:

y = a^x / log a

Assuming the log is to the base e:

y = (1/ log a) * a^x

Derivative of a^x:

Let u = a^x

log u = log a^x

log u = x log a

1/u * du/dx = log a

du/dx =  = u * log a

y = (1 / log a) * u

so  dy/du = 1/log a

and dy/dx = dy/du * du/dx

=  (1/log a )* log a u

= u

= a^x.