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  • Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Fill in the blanks Question 1

Provide Solution for RD Sharma Class 12 Chapter 11 Higher Order Derivatives Exercise Fill in the blanks Question 1

Answers (1)

Answer:  \frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}

Hint: Differentiate x & y w.r.t t. Use \frac{d y}{d x}=\frac{d y / d t}{d x / d t}. Again, differentiate w.r.t x.

Given:  y=t^{10}+1 \& x=t^{8}+1

Solution:

It is given that: y=t^{10}+1 \& x=t^{8}+1

Differentiating y w.r.t t,

\frac{d y}{d t}=10 t^{9}                                   …(i)                                \left[\because \frac{d}{d x} p^{n}=n p^{n-1}\right]

Differentiating x w.r.t t,

\frac{d x}{d t}=8 t^{7}                                   …(ii)

So, \frac{d y}{d x}=\frac{d y / d t}{d x / d t}

\begin{aligned} &\frac{d y}{d x}=\frac{10 t^{9}}{8 t^{7}} \\ & \end{aligned}

\frac{d y}{d x}=\frac{5}{4} t^{2}                               …(iii)

Again, differentiate (iii) w.r.t x:

\begin{aligned} &\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d x}\left(\frac{5}{4} t^{2}\right) \\ & \end{aligned}

\frac{d^{2} y}{d x^{2}}=\left(\frac{5}{4}\right) \frac{d}{d t}\left(t^{2}\right) \cdot \frac{d t}{d x}

\begin{aligned} &=\left(\frac{5}{4}\right)(2 t) \cdot \frac{1}{8 t^{7}} \\ & \end{aligned}                                                                         \left[\because \frac{d x}{d t}=8 t^{7}\right]

=\frac{10}{32 t^{6}}

Thus, \frac{d^{2} y}{d x^{2}}=\frac{5}{16 t^{6}}

 

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