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Please solve RD Sharma class 12 chapter 12 Derivative as a rate measure exercise Very short answer type question  9 maths textbook solution

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Answer: p^{\prime}(x)=30.255 \text { units }

Hint: Here, we use basic concept of rate change with respect to bodies and quantities

Given: p(x)=0.005 x^{3}+0.02 x^{2}+30 x

Solution:  Here

        p(x)=0.005 x^{3}+0.02 x^{2}+30 x

 Let’s differentiating with respect to x

        \begin{aligned} &p^{\prime}(x)=3(0.005) x^{2}+2(0.02) x+30 \\\\ &p^{\prime}(x)=0.015(x)^{2}+0.04 x+30 \\\\ &\text { Let } x=3 \end{aligned}


        \begin{aligned} &p^{\prime}(3)=0.015(3)^{2}+0.04(3)+30 \\\\ &=0.015(9)+0.12+30 \\\\ &=0.135+0.12+30 \\\\ &p^{\prime}(3)=30.255 \end{aligned}


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