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Q.
Let $y=t^{10}+1$ and $x=t^{8}+1$, then $\frac{d^{2} y}{d x^{2}}$ is
Continuity and Differentiability
Solution:
Here $y=t^{10}+1$ and $x=t^{8}+1$
$t^{8}=x-1 $
$\Rightarrow t^{2}=(x-1)^{1 / 4}$
So, $y=(x-1)^{5 / 4}+1$
Differentiate both sides w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{5}{4}(x-1)^{1 / 4}$
Again, differentiate both sides w.r.t. x, we get
$ \frac{ d ^{2} y }{ dx ^{2}}=\frac{5}{16}( x -1)^{-3 / 4} $
$\Rightarrow \frac{ d ^{2} y }{ dx ^{2}}=\frac{5}{16( x -1)^{3 / 4}}$
$=\frac{5}{16\left( t ^{2}\right)^{3}}=\frac{5}{16 t ^{6}}$