Question

If y = sec tan^-1x, then
dy/dx
is equal to

(a) xy/ (1 + x)
(b) xy/(1 + x²)^½
(c) y /(1 + x²)^½
(d) xy/(1 + x²)​

Answer 1

Answer:

correct answer is (d)xy/(1+x^2)

Step-by-step explanation:

given,y=sec (tan^-1x),

let, (tan^-1x)=a,

then,y=sec a,

& a=(tan^-1x) ,

also, tan a= x.

on differentiating 'y' with respect to 'a'.we get,

$\frac{dy}{da} = \frac{d \sec(a) }{da} = \sec(a) \tan(a)$

on differentiating 'a' with respect to 'x'.we get,

da/dx=d (tan^-1 x)/dx,

da/dx=1/(1+x^2).

Now,

$\frac{dy}{dx} = \frac{dy}{da} \times \frac{da}{dx} = sec(a) \tan(a) \times \frac{1}{1 + {x}^{2} } . \\ \frac{dy}{dx} = \frac{xy}{1 + {x}^{2} } .$

sec (a)= y & tan(a)=x.