# Differentiate:

Question:

Differentiate:.

$\left(x^{3} \cos x-2^{x} \tan x\right)$

Solution:

To find: Differentiation of $\left(x^{3} \cos x-2^{x} \tan x\right)$

Formula used: (i) (uv)' $=u^{\prime} v+u v^{\prime}$ (Leibnitz or product rule)

(ii)

$\frac{d x^{n}}{d x}=n x^{n-1}$

(iii)

$\frac{d \cos x}{d x}=-\sin x$

(iv)

$\frac{d a^{x}}{d x}=a^{x} \log a$

(v)

$\frac{d \tan x}{d x}=\sec ^{2} x$

Here we have two function $\left(x^{3} \cos x\right)$ and $\left(2^{x} \tan x\right)$

We have two differentiate them separately

Let us assume $g(x)=\left(x^{3} \cos x\right)$

And $h(x)=\left(2^{x} \tan x\right)$

Therefore, f(x) = g(x) – h(x)

$\Rightarrow f^{\prime}(x)=g^{\prime}(x)-h^{\prime}(x) \ldots$ (i)

Applying product rule on g(x)

Let us take $u=x^{3}$ and $v=\cos x$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(x^{3}\right)}{d x}=3 x^{2}$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\cos x)}{d x}=-\sin x$

Putting the above obtained values in the formula:-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left[x^{3} \cos x\right]^{\prime}=3 x^{2} \times \cos x+x^{3} \times-\sin x$

$=3 x^{2} \cos x-x^{3} \sin x$

$=x^{2}(3 \cos x-x \sin x)$

$g^{\prime}(x)=x^{2}(3 \cos x-x \sin x)$

Applying product rule on $\mathrm{h}(\mathrm{x})$

Let us take $u=2^{x}$ and $v=\tan x$

$u^{\prime}=\frac{d u}{d x}=\frac{d\left(2^{x}\right)}{d x}=2^{x} \log 2$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\tan x)}{d x}=\sec ^{2} x$

Putting the above obtained values in the formula:-

$(u v)^{\prime}=u^{\prime} v+u v^{\prime}$

$\left[2^{x} \tan x\right]^{\prime}=2^{x} \log 2 x \tan x+2^{x} \times \sec ^{2} x$

$=2^{x}\left(\log 2 \tan x+\sec ^{2} x\right)$

$h^{\prime}(x)=2^{x}\left(\log 2 \tan x+\sec ^{2} x\right)$

Putting the above obtained values in eqn. (i)

$f^{\prime}(x)=x^{2}(3 \cos x-x \sin x)-2^{x}\left(\log 2 \tan x+\sec ^{2} x\right)$

Ans) $x^{2}(3 \cos x-x \sin x)-2^{x}\left(\log 2 \tan x+\sec ^{2} x\right)$