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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 43 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 10 Differentiation Exercise 10.3 Question 43 Maths Textbook Solution.

Answers (1)

Answer: Hence Prove ,1+a^{2}=b

Hint:

\begin{aligned} &\frac{d}{d x}\left(x^{n}\right)=n x^{n-1} \\ &\frac{d(\text { constant) }}{d x}=0 \end{aligned}

Given:

\frac{d}{\mathrm{dx}}\left[\tan ^{-1}(\mathrm{a}+\mathrm{bx})\right]=1 \text { at } \mathrm{x}=0

Solution:

So, using chain rule,

we know \frac{d}{d x}\left(\tan ^{-1} x\right)=\frac{1}{1+x^{2}}

\begin{aligned} &{\left[\left\{\frac{1}{1+(a+b x)^{2}}\right\} \frac{d}{dx}(a+b x)\right]_{x=0}=1} \\ &{\left[\frac{1}{1+(a+b x)^{2}} \times(b)\right]_{x=0}=1} \\ &{\left[\frac{b}{\left[1+(a+b x)^{2}\right.}\right]_{x=0}=1} \\ &\therefore(a+b)^{2}=a^{2}+b^{2}+2 a b \end{aligned}

\begin{aligned} &{\left[\frac{b}{1+\left(a^{2}+b^{2} x^{2}+2 a b x\right)}\right]_{x=0}=1} \\ &\frac{b}{1+\left(a^{2}+0+0\right)}=1 \\ &\frac{b}{1+a^{2}}=1 \\ &b=1+a^{2} \end{aligned}

Hence Proved,

 

    

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