Question
If $\text{y}=\cos^{-1}\text{x},$ Find $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ in terms of y alone.
✓
Answer
Here,
$\text{y}=\cos^{-1}\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}^\frac{3}{2}}=\frac{-\text{x}}{(1-\text{x}^2)}$
Now,
$\text{y}=\cos^{-1}\text{x}$
$\Rightarrow\text{x}=\cos\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\cos\text{y}}{(1-\cos^2\text{y})^\frac{3}{2}}=-\frac{\cos\text{y}}{(\sin^2\text{y})^\frac{3}{2}}=-\cot\text{y}\ \text{cosec}^2\text{y}$
$\text{y}=\cos^{-1}\text{x},$
Differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{1}{\sqrt{1-\text{x}^2}}$
Differentiating w.r.t.x, we get
$\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-2\text{x}}{2\sqrt{1-\text{x}^2}^\frac{3}{2}}=\frac{-\text{x}}{(1-\text{x}^2)}$
Now,
$\text{y}=\cos^{-1}\text{x}$
$\Rightarrow\text{x}=\cos\text{y}$
$\Rightarrow\frac{\text{d}^2\text{y}}{\text{dx}^2}=\frac{-\cos\text{y}}{(1-\cos^2\text{y})^\frac{3}{2}}=-\frac{\cos\text{y}}{(\sin^2\text{y})^\frac{3}{2}}=-\cot\text{y}\ \text{cosec}^2\text{y}$
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