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Question Number 121727 by oustmuchiya@gmail.com last updated on 11/Nov/20

Answered by TANMAY PANACEA last updated on 11/Nov/20

$$1)b^2=\frac{a+1}{2}\to 2b\times \frac{db}{da}=\frac{1}{2}whena=4b=\sqrt{\frac{5}{2}}$$$$gradint{\left(\frac{db}{da}\right)}_{a=4}={\left(\frac{1}{4b}\right)}_{b=\sqrt{\frac{5}{2}}}=\frac{\sqrt{2}}{4\times \sqrt{5}}$$$$ii)\frac{dt}{dx}=\frac{x\times 2x-(x^2-1)\times 1}{x^2}=\frac{x^2+1}{x^2}=\frac{5}{4}$$$$iii)f=t^2-2t+3$$$$\frac{df}{dt}=2t-2=2\times 0.6-2=0.8$$$$iv)f\left(x\right)=\sqrt{x^3-1}$$$$\frac{df}{dx}=\frac{1}{2\sqrt{x^3-1}}\times 3x^2=\frac{3}{2}\times \frac{{(-2)}^2}{\sqrt{-8-2}}=\frac{6}{\sqrt{-10}}\Lleftarrow lookhere\sqrt{-10}iscoming$$$$$$

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