# If f(x) =tan^24x  and g(x) = sqrt(5x-1 , what is f'(g(x)) ?

Jul 3, 2016

$f ' \left(g \left(x\right)\right) = 24 {\tan}^{23} \left(\sqrt{5 x - 1}\right) \times {\sec}^{2} \left(\sqrt{5 x - 1}\right)$

#### Explanation:

As $f \left(x\right) = {\tan}^{24} x$ and $g \left(x\right) = \sqrt{5 x - 1}$

As $f \left(x\right) = {\tan}^{24} x$

$\frac{\mathrm{df}}{\mathrm{dx}} = f ' \left(x\right) = 24 {\tan}^{23} x \times {\sec}^{2} x$

and $f ' \left(g \left(x\right)\right) = 24 {\tan}^{23} \left(\sqrt{5 x - 1}\right) \times {\sec}^{2} \left(\sqrt{5 x - 1}\right)$

Jul 4, 2016

Reqd. Deri. $= \left(\frac{48}{5}\right) {\tan}^{23} x \cdot {\sec}^{2} x \cdot \sqrt{5 x - 1} .$

#### Explanation:

$f \left(x\right) = {\tan}^{24} x , g \left(x\right) = \sqrt{5 x - 1} = t ,$ say.

Now Reqd. Deri. $= f ' \left(g \left(x\right)\right) = f ' \left(t\right) = \frac{d}{\mathrm{dt}} \left\{f \left(t\right)\right\} = \frac{\mathrm{df}}{\mathrm{dt}}$

We see that $f$ is a fun. of $x$, and $x$ of $t$. Hence,

reqd. Deri. $= \frac{\mathrm{df}}{\mathrm{dt}} = \frac{\mathrm{df}}{\mathrm{dx}} \cdot \frac{\mathrm{dx}}{\mathrm{dt}} \ldots \ldots \ldots . \left(1\right)$

Now, $f \left(x\right) = {\tan}^{24} x \Rightarrow \frac{\mathrm{df}}{\mathrm{dx}} = 24 {\tan}^{23} x \cdot \frac{d}{\mathrm{dx}} \tan x = 24 {\tan}^{23} x \cdot {\sec}^{2} x$.....(2)

Next, t=sqrt(5x-1) rArr dt/dx=1/(2sqrt(5x-1))*d/dx(5x-1)=5/(2sqrt(5x-1) $\Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}} = \frac{2 \sqrt{5 x - 1}}{5.} \ldots \ldots \ldots \ldots \ldots \ldots . \left(3\right)$

From $\left(1\right) , \left(2\right) , \left(3\right)$, Reqd. Deri. $= 24 {\tan}^{23} x \cdot {\sec}^{2} x \cdot \frac{2 \sqrt{5 x - 1}}{5}$,
$= \left(\frac{48}{5}\right) {\tan}^{23} x \cdot {\sec}^{2} x \cdot \sqrt{5 x - 1} .$