If `f^(prime)(x)=sqrt(2x^2-1)` and `y=f(x^2)` , then find `(dy)/(dx)` at `x=1` .

Explanation

`text { We have, } f^{prime}(x)=sqrt{2 x^{2}-1} `

`text { and } y=f(x^{2}) `

`Rightarrow frac{d y}{d x}=frac{d}{d x} f(x^{2}) `

`Rightarrow frac{d y}{d x}=f^{prime}(x^{2}) frac{d}{d x}(x^{2}) `

`Rightarrow frac{d y}{d x}=f^{prime}(x^{2}) times 2 x `

`Rightarrow frac{d y}{d x}=2 x f^{prime}(x^{2}) `

`text { Putting } x=1, text { we get } `

`frac{d y}{d x}=2(1) f^{prime}(1^{2}) `

...