Question:

If $2^x+2^y = 2^{x+y}$, then $\frac {dy}{dx}$ is

Updated On: Jul 14, 2024
  • $2^{y-x}$
  • $-2^{y-x}$
  • $2^{x-y}$
  • $\frac {2^y-1}{2^x-1}$
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The Correct Option is B

Solution and Explanation

$2^{x}+2^{y}=2^{(x+y)} \ldots . .(1)$
Differentiating both sides w.r.t. $x$, we get
$2^{x} \ln 2+2^{y} y' \ln 2=2^{(x+y)} \ln 2\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} y'=2^{(x+y)}\left(1+y'\right)$
$\Rightarrow 2^{x}+2^{y} \cdot y'=2^{(x+y)}+2^{(x+y)} \cdot y'$
$\Rightarrow 2^{x}-2^{(x+y)}=y'\left(2^{(x+y)}-2^{y}\right)$
$\Rightarrow 2^{x}-2^{x}-2^{y}=y'\left(2^{x}+2^{y}-2^{y}\right)[$ From eqn. $(1)]$
$\Rightarrow-2^{y}=y'2^{x}$
$\Rightarrow y'=d y / d x=-2^{y-x}$
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Concepts Used:

Differentiability

Differentiability of a function A function f(x) is said to be differentiable at a point of its domain if it has a finite derivative at that point. Thus f(x) is differentiable at x = a
\(\frac{d y}{d x}=\lim _{h \rightarrow 0} \frac{f(a-h)-f(a)}{-h}=\lim _{h \rightarrow 0} \frac{f(a+h)-f(a)}{h}\)
⇒ f'(a – 0) = f'(a + 0)
⇒ left-hand derivative = right-hand derivative.
Thus function f is said to be differentiable if left hand derivative & right hand derivative both exist finitely and are equal.
If f(x) is differentiable then its graph must be smooth i.e. there should be no break or corner.
Note:
(i) Every differentiable function is necessarily continuous but every continuous function is not necessarily differentiable i.e. Differentiability ⇒ continuity but continuity ⇏ differentiability

(ii) For any curve y = f(x), if at any point \(\frac{d y}{d x}\) = 0 or does not exist then, the point is called “critical point”.

3. Differentiability in an interval
(a) A function fx) is said to be differentiable in an open interval (a, b), if it is differentiable at every point of the interval.

(b) A function f(x) is differentiable in a closed interval [a, b] if it is

  • Differentiable at every point of interval (a, b)
  • Right derivative exists at x = a
  • Left derivative exists at x = b.