Question

If tan (x + y) = x, then dy/dx at (0, 0) is :
(a) 1 (b) -1 (c) 0 (d) 2
need answer with explanation​

Answer 1

Answer:

→ tan(x + y) = x.

Then dy/dx at (0,0).

As we know that,

Differentiate both sides w.r.t x, we get.

d[tan(x + y)]/dx = d[x]/dx.

⇒ sec²(x + y). d[x+y]/dx = 1.

→ sec²(x + y).(1 + dy/dx) = 1. sec²(x + y) + sec²(x + y)dy/dx = 1.

sec²(x + y)dy/dx = 1 - sec²(x + y).

➡dy/dx = [1 - sec²(x + y)]/[sec²(x + y)].

Put the value of x = 0 and y = 0 in the equation, we get.

→ dy/dx = [1 - sec²(0 + 0)]/[sec²(0 + 0)].

→ dy/dx = [1 - 1]/1.

⇒dy/dx = 0.

Option [C] is correct answer.

MORE INFORMATION.

(1) d(sin x)/dx = cos X.

(2) d(cos x)/dx = - sin x.

(3) d(tan x)/dx = sec²x.

(4) d(cot x)/dx = - cosec²x.

(5) d(sec x)/dx = sec x tan x.

(6) d(cosec x)/dx = - cosec x cot x.

Answer 2

Answer:

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