You visited us 1 times! Enjoying our articles? Unlock Full Access!

First Principle of Differentiation

Differentiate...

Question

Differentiate from first principle:
$(x i i) 3^{x^{2}}$

Open in App

Solution

Given:
$f (x) = 3^{x^{2}}$

The derivative of a function $f (x)$ is defined as:

$f^{'} (x) = lim h \to 0 \frac{f (x + h) - f (x)}{h}$

Putting $f (x)$ in above expression, we get:
$\Rightarrow f^{'} (x) = lim h \to 0 \frac{3^{{(x + h)}^{2}} - 3^{x^{2}}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{3^{x^{2}} (3^{{(x + h)}^{2} - x^{2}} - 1)}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{3^{x^{2}} (3^{{(x + h)}^{2} - x^{2}} - 1)}{{(x + h)}^{2} - x^{2}} \times \frac{{(x + h)}^{2} - x^{2}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{3^{x^{2}} (3^{{(x + h)}^{2} - x^{2}} - 1)}{{(x + h)}^{2} - x^{2}} \times \frac{2 h x + h^{2}}{h}$

$\Rightarrow f^{'} (x) = lim h \to 0 \frac{3^{x^{2}} (3^{{(x + h)}^{2} - x^{2}} - 1)}{{(x + h)}^{2} - x^{2}} \times (2 x + h)$

$[∵ lim x \to 0 \frac{a^{x} - 1}{x} = {log}_{e} a]$

$\Rightarrow f^{'} (x) = 3^{x^{2}} \times l o g_{e} 3 \times (2 x + 0)$

$\Rightarrow f^{'} (x) = 2 x (3^{x^{2}}) l o g_{e} 3$

Therefore, the derivative of $3^{x^{2}}$ is $2 x (3^{x^{2}}) l o g_{e} 3$

Suggest Corrections

Similar questions

\sqrt{\sin 2 x}

\frac{\sin x}{x}

\frac{\cos x}{x}

\sqrt{\sin (3 x + 1)}

e^{x^{2} + 1}

e^{\sqrt{2 x}}

e^{\sqrt{a x + b}}

a^{\sqrt{x}}

3^{x^{2}}

Differentiate each the following from first principles :

$(i) e^{- x}$

$(i i) e^{3 x}$

$(i i i) e^{a x + b}$

$(i v) x e^{x}$

$(v) x^{2} e^{x}$

$(v i) e^{x^{2} + 1}$

$(v i i) e^{\sqrt{2 x}}$

$(v i i i) e^{\sqrt{a x + b}}$

$(i x) a^{\sqrt{x}}$

$(x) 3^{x^{2}}$

\sqrt{\frac{(x - 3) (x^{2} + 4)}{3 x^{2} + 4 x + 5}}

\sqrt{\frac{(x - 3) (x^{2} + 4)}{3 x^{2} + 4 x + 5}}

x

- 3 x^{2} . (sin 2 x^{3}) {cos [{cos}^{2} (x^{3})]}

Join BYJU'S Learning Program

Explore more

Join BYJU'S Learning Program