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Question Number 133616 by benjo_mathlover last updated on 23/Feb/21

$$\mathrm{Suppose}\mathrm{function}\mathrm{f}:\mathrm{R}\to \mathrm{R}\mathrm{with}$$$$\mathrm{f}(2\mathrm{x}-3)={4\mathrm{x}}^2+2\mathrm{x}-5\mathrm{and}\mathrm{f}{}^{\prime }\mathrm{denote}$$$$\mathrm{is}\mathrm{derivatif}\mathrm{of}\mathrm{f}\mathrm{function}.$$$$\mathrm{What}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{f}{}^{\prime }(2\mathrm{x}-3)$$

Commented by mr W last updated on 23/Feb/21

$$4x+1$$

Answered by TheSupreme last updated on 23/Feb/21

$$D\left(f\left(g\left(x\right)\right)\right)=g^{\prime }\left(x\right)f^{\prime }\left(g\left(x\right)\right)=8x+2$$$$g\left(x\right)=2x-3\to g^{\prime }\left(x\right)=2$$$$f^{\prime }\left(g\left(x\right)\right))=4x+1$$

Answered by liberty last updated on 23/Feb/21

$$\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=4{\left(\frac{\mathrm{x}+3}{2}\right)}^2+2\left(\frac{\mathrm{x}+3}{2}\right)-5$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+6\mathrm{x}+9+\mathrm{x}+3-5$$$$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{x}}^2+7\mathrm{x}+7$$$$\Rightarrow \mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=2\mathrm{x}+7$$$$\mathrm{f}{}^{\prime }(2\mathrm{x}-3)=2(2\mathrm{x}-3)+7=4\mathrm{x}+1$$

Answered by mr W last updated on 23/Feb/21

$$letu=2x-3$$$$\Rightarrow \frac{dx}{du}=\frac{1}{2}$$$$\mathrm{f}\left(u\right)={4\mathrm{x}}^2+2\mathrm{x}-5$$$$f^{\prime }\left(u\right)=\frac{df}{du}=\frac{df}{dx}\times \frac{dx}{du}=(8x+2)\frac{1}{2}=4x+1$$$$i.e.f^{\prime }(2x-3)=4x+1$$

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