Explanation We can use here the formula for derivative of sin−1x, which is d dx sin−1x = 1 √1 − x2 As such to find derivative dy dx for y = sin−12x using chain rule is given by dy dx = 1 √1 − (2x)2 × d dx (2x) = 2 √1 −4x2 Answer link Explanation siny = x2 cosy( dy dx) = 2x dy dx = 2x cosy dy dx = 2x √1 − sin2y dy dx = 2x √1 − (x2)2 dy dx = 2x √1 −x4 Hopefully this helps!Simplify where possibley = sin−1(2x 1) This problem has been solved!
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1 quad 9.y=sin^(-1)((2x)/(1+x^(2)))
1 quad 9.y=sin^(-1)((2x)/(1+x^(2)))-Find Dy Dx If Y Sin 1 2x 1 X 2 hviezdoslavov kubín 7 ročník hugolin gavlovic valaska skola mravov stodola if f x x 3 x 2 x 1 then f 2 humanizmus a renesancia znaky ii rákóczi ferenc portrSolution Solution y = sin − 1 (1 x 2 2 x ) ⇒ sin y = 1 x 2 2 x ⇒ cos y = 1 − sin 2 y = 1 − (1 x 2 2 x ) 2 = 1 x 2 1 − x 2 Differentiating it wrt x, cos y d x d y = (1 x 2) 2 2 (1 x 2) − 2 x (2 x) ⇒ cos y d x d y = (1 x 2) 2 2 (1 − x 2) ⇒ d x d y = cos y 1 ((1 x 2) 2 2 (1 − x 2) ) ⇒ d x d y = 1 − x 2 1 x 2 ∗ ((1 x 2) 2 2 (1 − x 2) ) ⇒ d x d y = 1 x 2 2




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If y = (tan1 x) 2, prove that ( 1 x 2) 2 y 2 2x( 1Sin (θ), Tan (θ), and 1 are the heights to the line starting from the x axis, while Cos (θ), 1, and Cot (θ) are lengths along the x axis starting from the origin The functions sine, cosine and tangent of an angle are sometimes referred to as the primary or basic trigonometric functionsIf \y = \sin^{ 1} \left( \frac{2x}{1 x^2} \right) \sec^{ 1} \left( \frac{1 x^2}{1 x^2} \right), 0 < x < 1,\ prove that \\frac{dy}{dx} = \frac{4}{1 x^2}\ ?
If `y=sin^(1)((x^21)/(x^21))sec^(1)((x^21)/(x^21))` then `dy/dx` is equal toIf y = sin (1x^2)/(1x^2), then dy/dx = Login Study Materials NCERT Solutions NCERT Solutions For Class 12 NCERT Solutions For Class 12 Physics;Y=3tan−1 x− √ x2 1 ⇒ y′ = 3 1 x− √ x2 1 2 1− x √ x2 1 ⇒ y′ = 3 1x 2−2x √ x2 1x 1 √ x2 1−x √ x2 1 ⇒ y′ = 3(√ x2 1−x



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Best answer Let y = sin1( (2x 1)/ (1 4x)) = sin1( (2x x 2)/ (1 (2x)2)) Let 2x = tanθ ⇒ θ = tan12x then we have y = sin1(2tanθ/ (1 tan2θ)) = sin1(sin2θ) = 2θ = 2tan12x Differentiating both sides wrt x, we get Please log in or register to add a comment ← Prev Question Next Question → If y= sin^(1)((2x)/(1x^2)) sec^(1)((1x^2)/(1x^2)) , prove that (dy)/(dx)=4/(1x^2)Find dy/dx of ,y=1/sin(2x/1x^2) Ask questions, doubts, problems and we will help you menu myCBSEguide Courses CBSE Entrance Exam Competitive Exams ICSE & ISC Teacher Exams UP Board Uttarakhand Board Features Online Test Practice Homework Help Downloads CBSE Videos



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Ex 53, 9 Find 𝑑𝑦/𝑑𝑥 in, y = sin^(−1) (2𝑥/( 1 2𝑥2 )) 𝑦 = sin^(−1) (2𝑥/( 1 2𝑥2 )) Putting x = tan θ 𝑦 = sin^(−1) (2𝑥/( 1COMEDK 08 If y = sin1 ((5x12 √1 x2/13)) , then (dy/dx) = (A) (3/√1 x2) (B) (12/√1 x2) (1/√1 x2) (D) (1/√1 x2) Check A NCERT Solutions for Class 12 Maths Chapter 5 Continuity and Differentiability Ex 53 Find in the following Ex 53 Class 12 Maths Question 1 2x 3y = sinx Solution 2x 3y = sinx Differentiating wrt x, => Ex 53 Class 12 Maths Question 2




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Consider the triangle T ⊂ S with vertices (0,0), (1/2,1/2), (1/2,1) Thus, T is defined by the inequalities 0 < x < y < 2x < 1 For every (x,y) in T, xy > x2 and x2 y2 < 5x2 Show that all solutions of y'= \frac {xy1} {x^21} are of the form y=xC\sqrt {1x^2} without solving the ODE Show that all solutions of y′ = x21xy1 sin1 ( x root (1x) root(x)root(1x^2) ) pls answer , I have a test tomorrow / Share with your friends Share 11 take x = sin a & rootx = sin b then it will be sin^1 (sin a root (1 sin^2b) sin b root (1sin^2a) ) sin^1 (sina cosb sinb cos a) sin^1 (sin (ab)) a b sin^1x sin^1rootx 18 ;Let y = sin1 ((2x/1x2)), 0 < x < 1 and 0 < y < (π/2), then (dy/dx) is equal to Q Let $y = \sin^{1} \left(\frac{2x}{1x^{2}}\right), 0 x 1$ and $ 0 y \frac{\pi}{2}, $ then $\frac{dy}{dx} $ is equal to




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X 2 Y 2 Z 2 16;2x (1x 2)(2x)/(1x 2) 2 = cos (1x 2)/(1x 2) 2x 2x 3 2x2x 3/(1x 2) 2 = cos (1x 2)/(1x 2) 4x/(1x 2) 2 = (4x/(1x 2) 2 ) cos ((1x 2)/(1x 2)) Hence option (3) is the answer Was Ex 53, 14 Find 𝑑𝑦/𝑑𝑥 in, y = sin–1 (2𝑥 √ (1−𝑥^2 )) , − 1/√2 < x < 1/√2 y = sin–1 (2𝑥 √ (1−𝑥^2 )) Putting 𝑥 =𝑠𝑖𝑛𝜃 𝑦 = sin–1 (2 sin𝜃 √ (1−〖𝑠𝑖𝑛〗^2 𝜃)) 𝑦 = sin–1 ( 2 sin θ √ (〖𝑐𝑜𝑠〗^2 𝜃)) 𝑦 ="sin–1 " (〖"2 sin θ" 〗cos𝜃 ) 𝑦 = sin–1 (sin〖2 𝜃)〗 𝑦 = 2θ Putting value of θ = sin−1 x 𝑦 = 2 〖𝑠𝑖𝑛〗^ (−1) 𝑥 Since x = sin θ ∴




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So,the 'y' in the question,arcsin(2x/1x^2) is a little difficult to handle,so a smart substitution has been done in the form of x=tan θ which simplifies the 'y' to be equal to 2 arctan(x) Now,y=2tan^1(x) Differentiating both sides,we get dy/dx=2*1/1x^2 as derivative of tan^1(x) is 1/1x^2 And dy/dx is what was asked in the question Cheers )




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