Question

If $u={\sin }^{-1}(x-y);x=3t;y=4t^3$, then $\frac{du}{dt}$ is equal to

• A. $\frac{1}{\sqrt{1-t^2}}$
• B. $\frac{2}{\sqrt{1-t^2}}$
• C. $\frac{3}{\sqrt{1-t^2}}$
• D. $\frac{4}{\sqrt{1-t^2}}$

Answer 1

Answer: (C)

$\frac{3}{\sqrt{1-t^2}}$

The given information is: $u={\sin }^{-1}(x-y)$, $x=3t$ and $y=4t^3$

Substituting the values of $x$ and $y$ in the function $u$ we get,

$\Rightarrow u={\sin }^{-1}(3t-4t^3)$

Differentiating with respect to $t$ we get,

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{1-(3t-4t^3{)}^2}}$

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{1-t^2(3-4t^2{)}^2}}$

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{1-t^2(9+16t^4-24t^2)}}$

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{1-9t^2-16t^6+24t^4}}$

Let us take $t^2=z$ so that the quation changes to a cubic equation.

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{1-9z-16z^3+24z^2}}$

$\Rightarrow \frac{du}{dt}=\frac{3-12t^2}{\sqrt{-(z-1)(4z-1{)}^2}}$

$\Rightarrow \frac{du}{dt}=\frac{3(4z-1)}{\sqrt{1-z}(4z-1)}$

$\Rightarrow \frac{du}{dt}=\frac{3}{\sqrt{1-z}}$

$\Rightarrow \frac{du}{dt}=\frac{3}{\sqrt{1-t^2}}$