Question

how to differentiate it step by step​

Answer 1

Answer:

$- 20 (1-2x)^{2/3} - 3cosec^2(x)$

Step-by-step explanation:

$y = 3 cot (x) + 6(1-2x)^{5/3}$

$y=u+v$

$\frac{dy}{dx} =\frac{du}{dx} +\frac{dv}{dx}$     ( 1 )

$u = 3 [cot x]$

$\frac{du}{dx} = 3[\frac{d}{dx} (cotx) ]$

Differentiation of $cot x = -cosec^2 x$:

$\frac{du}{dx} = 3[-cosec^2x]$

$\frac{du}{dx} = -3cosec^2(x)$

$v = 6 [1-2x]^{5/3}$

$\frac{dv}{dx} = 6[\frac{d}{dx} (1 - 2x)^{5/3}]$

Differentiation of $x^n = n \times x ^ {(n-1)}$:

$\frac{dv}{dx} = 6[\frac{5}{3} \times (1 - 2x)^{(\frac{5}{3} -1)} \times \frac{d}{dx}(1-2x) ]$

Differentiation of a constant is 0:

$\frac{dv}{dx} = 6[\frac{5}{3} \times (1 - 2x)^{(\frac{5}{3} -1)} \times (0-2) ]$

Taking out all constants:

$\frac{dv}{dx} = [\frac{6 \times 5 \times -2}{3} \times (1 - 2x)^{(\frac{5-3}{3})} ]$

$\frac{dv}{dx} = [-20 \times (1 - 2x)^{(\frac{2}{3})} ]$

Substituting in ( 1 ):

$\frac{dy}{dx} = [-3cosec^2(x)] + [-20 \times (1 - 2x)^{(\frac{2}{3})}]$

You can solve both terms together as well!

Any corrections or suggestions are welcome :)