If y=∫ limitsu(x)v(x) f(t) d t, let us define (d y/d x) in a different manner as (d y/d x)=v prime(x) f2(v(x))-u prime(x) f2(u(x)) and the equation of the tangent at (a, b) as y-b=((d y/d x))(a, b)(x-a) If y=∫ limitsxx2 t2 d t, then equation of tangent at x=1 is

Question

If y=∫ limitsu(x)v(x) f(t) d t, let us define (d y/d x) in a different manner as (d y/d x)=v prime(x) f2(v(x))-u prime(x) f2(u(x)) and the equation of the tangent at (a, b) as y-b=((d y/d x))(a, b)(x-a) If y=∫ limitsxx2 t2 d t, then equation of tangent at x=1 is (A) y=x+1 (B) x+y=1 (C)  y=x-1 (D) y=x

Tardigrade Admin · Accepted Answer

At x=1, y=0 (d y/d x)=2 x ⋅ x8-x4=2-1=1 ∴ equation of tangent y-0=1(x-1) y=x-1

$y = x + 1$

$x + y = 1$

$y = x - 1$

$y = x$

At $x = 1, y = 0$
$\frac{d y}{d x} = 2 x \cdot x^{8} - x^{4} = 2 - 1 = 1$
$∴$ equation of tangent $y - 0 = 1 (x - 1)$
$y = x - 1$

Q. If y=u(x)∫v(x)​f(t)dt, let us define dxdy​ in a different manner as dxdy​=v′(x)f2(v(x))−u′(x)f2(u(x)) and the equation of the tangent at (a,b) as y−b=(dxdy​)(a,b)​(x−a) If y=x∫x2​t2dt, then equation of tangent at x=1 is

y=x+1

x+y=1

y=x−1

y=x

At x=1,y=0 dxdy​=2x⋅x8−x4=2−1=1 ∴ equation of tangent y−0=1(x−1) y=x−1

$y = x + 1$

$x + y = 1$

$y = x - 1$

$y = x$

At $x = 1, y = 0$
$\frac{d y}{d x} = 2 x \cdot x^{8} - x^{4} = 2 - 1 = 1$
$∴$ equation of tangent $y - 0 = 1 (x - 1)$
$y = x - 1$