Question

Can someone please differentiate this?

Answer 1

Answer:

b. cos2x(sin2x)^-1/2 is the answer

Explanation:

Apply chain rule, first differentiate root part then sin2x part then 2x part

after differentiation,

√sin2x becomes 1/2√sin2x

sin2x becomes cos2x

2x becomes 2

multiply all,

you know that √z = z^1/2 and 1/√z = z^-1/2

similarly 1/√sin2x = (sin2x)^-1/2

Answer 2

Solution :

$\sf{\dfrac{dy}{dx} = \sqrt{sin(2x)}}$

$\textsf{By applying the chain rule of differentiation}$$\textsf{and substituting the values in it, we get :}$

$\boxed{\begin{minipage}{7 cm} \underline{\sf{Chain\:rule\:of\:differentiation :-}} \\ \\ \sf{\dfrac{dx}{dy} = \dfrac{dy}{du} \cdot \dfrac{du}{dx}} \end{minipage}}$

$\sf{Here,} \\ \\ \bullet \textsf{Function of y} \sf{= \sqrt{sin(2x)}} \\ \bullet \textsf{Function of u} \sf{= sin(2x)}$

$:\implies \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{d[\sqrt{sin(2x)}]}{d[sin(2x)]} \times \dfrac{d[sin(2x)]}{dx}}$

$:\implies \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{d[sin(2x)]^{1/2}}{d[sin(2x)]} \times \dfrac{d[sin(2x)]}{dx}}$

$\underline{\sf{Differentiate\:of\:[sin(2x)]^{1/2}]}} :$

$\textsf{By applying the power rule of differentiation and substituting the values in it, we get :}$

$\boxed{\begin{minipage}{7 cm} \underline{\sf{Power\:rule\:of\:differentiation :-}} \\ \\ \sf{\dfrac{d(x^{n})}{dy} = n \cdot x^{(n - 1)}} \end{minipage}}$

$:\implies \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{1}{2} \times sin(2x)^{(1/2 - 1)}}$

$:\implies \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{1}{2} \times sin(2x)^{(1/2 - 1)}}$

$:\implies \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{1}{2} \times sin(2x)^{(1 - 2)/2}}$

$:\implies \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{sin(2x)^{-1/2}}{2}}$

$:\implies \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{1}{2\sqrt{sin(2x)}}}$

$\boxed{\therefore \sf{\dfrac{d[sin(2x)]^{1/2}]}{dy} = \dfrac{1}{2\sqrt{sin(2x)}}}}$

$\underline{\sf{Differentiate\:of\:[sin(2x)]]}} :$

$:\implies \sf{\dfrac{d[sin(2x)]}{dy} = \dfrac{d[sin(2x)]}{dx}}$

$\textsf{By applying the chain rule of differentiation}$$\textsf{ and substituting the values in it, we get :}$

$\sf{Here,} \\ \\ \bullet \textsf{Function of y} \sf{= sin(2x)} \\ \bullet \textsf{Function of u} \sf{= 2x}$

$:\implies \sf{\dfrac{d[sin(2x)]}{dy} = \dfrac{d[sin(2x)]}{d(2x)} \times \dfrac{d(2x)}{dx}}$

$:\implies \sf{\dfrac{d[sin(2x)]}{dy} = cos(2x) \times 2} \:\:\:\:[\because \sf{\dfrac{d[sin(x)]}{dx} = cos(x)}]$

$:\implies \sf{\dfrac{d[sin(2x)]}{dy} = 2cos(2x)}$

$\boxed{\therefore \sf{\dfrac{d[sin(2x)]}{dy} = 2cos(2x)}}$

$\textsf{Now, by substituting the derivative of}$$\: \sf{[sin(2x)]^{1/2}\:and\:[sin(2x)]} \: \textsf{in the equation, we get :}$

$:\implies \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{d[sin(2x)]^{1/2}}{d[sin(2x)]} \times \dfrac{d[sin(2x)]}{dx}}$

$:\implies \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{1}{2\sqrt{sin(2x)}} \times 2cos(2x)}$

$:\implies \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{2cos(2x)}{2\sqrt{sin(2x)}}}$

$\boxed{\therefore \sf{\dfrac{d[\sqrt{sin(2x)}]}{dy} = \dfrac{2cos(2x)}{2\sqrt{sin(2x)}}}}$