Differentiate the following w.r.t x
(i) $y=\sqrt{sin x^3}$
$\frac{dy}{dx}=\frac{1}{sin x^3} \frac{d}{dx}(sin x^3)$
$=\frac{1}{2\sqrt{sin x^3}} cos x^3\frac{d}{dx} x^3$
$=\frac{cos x^3}{sin x^3}3x^2 $
$=\frac{3x^2 cos x^3}{sin x^3}$
(ii)`y=log[cos(x^5)]`
`\frac{dy}{dx}=\frac{d}{dx}(log[cos(x^5)])`
`=\frac{1}{cos(x^5)}\frac{d}{dx}[cos(x^5)]`
`=\frac{1}{cos(x^5)}-sinx^5\frac{d}{dx} x^5`
`=\frac{-sin x^5}{cos x^5}5x^3`
`=\frac{-5x^3}{cos x^5}`
(iii)`y=cot^{2}(x^3)`
`\frac{dy}{dx}=2cot(x^3)\frac{d}{dx}[cot(x^3)]`
`=2cot(x^3)-cosec x^3 \frac{d}{dx} x^3`
`=-2cot(x^3)cosec^2(x^3) 3x^2`
`=-6x^2cot(x^3)cosec^2(x^3)`
(iv)`y=log[cos(x^5)]`
`\frac{dy}{dx}=\frac{1}{cos(x^5)}\times\frac{d}{dx}cos(x^5)`
`=\frac{1}{cos(x^5)}\times -sin(x^5)\frac{d}{dx}x^5`
`= \frac{ -sin(x^5)}{cos(x^5)}5x^4`
`= \frac{ -5x^4sin(x^5)}{cos(x^5)}`
(v)`y=(x^3+2x-3)^4(x+cos x)^3`
`\frac{dy}{dx}=4(x^3+2x-3)^3\frac{d}{dx}x^3+2x-3 \times 3(x+cos x)^2\frac{d}{dx}x+cos x`
`=4(x^3+2x-3)^3\left(3x^2+2\right) \times 3(x+cos x)^2 (1-sin x)`
(vi)`y=(1+cos^2 x)^4 \times \sqrt{x+\sqrt{tan x}}`
`\frac{dy}{dx}=\frac{d}{dx}(1+cos^2 x)^4`` \times \sqrt{x+\sqrt{tan x}}`
`=(1+cos^2 x)^4 ``\frac{d}{dx}\sqrt{x+\sqrt{tan x}}``+ \sqrt{x+\sqrt{tan x}}``\frac{d}{dx}(1+cos^2 x)^4`
`=(1+cos^2 x)^4`` \frac{1}{2\sqrt{x+\sqrt{tan x}}}``\frac{d}{dx}(x+\sqrt{tan x})``+ \sqrt{x+\sqrt{tan x}}``\times`` 4 (1+cos^2 x)^3``\frac{d}{dx}(1+cos^2 x)`
`=(1+cos^2 x)^4`` \frac{1}{2\sqrt{x+\sqrt{tan x}}}``\left(1 + \frac{1}{2\sqrt{tan x}}\right)``\frac{d}{dx}tanx``+ \sqrt{x+\sqrt{tan x}}``\times``4 (1+cos^2 x)^3\times`` \frac{d}{dx}(1+cos^2 x)`
`=(1+cos^2 x)^4 ``\frac{1}{2\sqrt{x+\sqrt{tan x}}``\left(1 + \frac{1}{2\sqrt{tan x}}\right) ``sec^2x \sqrt{x+\sqrt{tan x}}``\times``4 (1+cos^2 x)^3``\times``2cos x``\frac{d}{dx}cos x`
`=(1+cos^2 x)^4 ``\frac{1}{2\sqrt{x+\sqrt{tan x}}``\times``1 + \frac{1}{2\sqrt{tan x}`` sec^2x``\sqrt{x+\sqrt{tan x}}``\times 4``(1+cos^2 x)^3``\times``-2cos x sinx`
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| Differentiate the following w.r.t x |
