Download App

Differentiation — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsDifferentiation1 Mark
MCQ
If $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y}),$ then $\frac{\text{dy}}{\text{dx}}=$
  • A
    2
  • B
    -2
  • C
    1
  • ✓
    -1
✓

Answer

Correct option: D.
-1
We have, $\sin(\text{x}+\text{y})=\log(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\Big(1+\frac{\text{dy}}{\text{dx}}\Big)=\frac{1}{(\text{x}+\text{y})}\Big(1+\frac{\text{dy}}{\text{dx}}\Big)$
$\Rightarrow\cos(\text{x}+\text{y})+\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}+\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{x}+\text{y}}-\cos(\text{x}+\text{y})$
$\Rightarrow\cos(\text{x}+\text{y})\frac{\text{dy}}{\text{dx}}-\frac{1}{(\text{x}+\text{y})}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\cos(\text{x}+\text{y})-\frac{1}{(\text{x}+\text{y})}\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\Big\{\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})\Big\}\frac{\text{dy}}{\text{dx}}=\frac{1}{(\text{x}+\text{y})}-\cos(\text{x}+\text{y})$
$\Rightarrow\frac{\text{dy}}{\text{dx}}=-1$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Differentiation→Differentiation — all question types→Maths STD 12 Science question papers→Gujarat Board STD 12 Science question papers→Gujarat Board question paper generator→

Similar questions

If $x({x^4} + 1)\phi (x) = 1,$ then $\int_1^2 {\phi (x)\,dx = } $
→
The area of the region bounded by the curves $y = {x^2}$ and $y = \,|x|$ is
→
The direction ratios of the line x - y + z - 5 = 0 = x - 3y - 6 are proportional to:
→
If $y=\log _7(\log x)$, then find $\frac{d y}{d x}$.
→
If $f(x) = 2{x^3} - 3{x^2} - 12x + 5$ and $x \in [ - 2,\,4]$, then the maximum value of function is at the following value of $x$
→
Let the system of equations $x+2 y+3 z=5$, $2 x+3 y+z=9,4 x+3 y+\lambda z=\mu$ have infinite number of solutions. Then $\lambda+2 \mu$ is equal to :
→
$a = 3i - 5j$ and $b = 6i + 3j$ are two vectors and $c$ is a vector such that $c = a \times b$, then $|a|:|b|:|c|$ is
→
Two events $A$ and $B$ will be independent, if
→
Let $f\,:\,R \to R$ be a function such that $f\left( x \right) = {x^3} + {x^2}f'\left( 1 \right) + xf''\left( 2 \right) + f'''\left( 3 \right)$, $x \in R$. Then $f(2)$ equals
→
The region represented by the inequation system x, y ≥ 0, y ≤ 6, x + y ≤ 3 is:
→