not sure how todo differentiations can any one explain how to do this equasion y=3x^4 -3√x+8
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Originally Posted by bart 1000 not sure how todo differentiations can any one explain how to do this equasion y=3x^4 -3√x+8 $\displaystyle y = 3x^4 - 3x^{1/2} + 8$ $\displaystyle \frac{dy}{dx} = 12x^3 - 1.5x^{-1/2}$
Originally Posted by bart 1000 not sure how todo differentiations can any one explain how to do this equasion y=3x^4 -3√x+8 First lets quote the power rule $\displaystyle \frac{d}{dx}x^n=nx^{n-1}$ Now lets rewrite y with fractional exponents using $\displaystyle \sqrt[b]{x^a}=x^{\frac{a}{b}}$ So $\displaystyle y=3x^4-3x^{\frac{1}{2}}+8$ Just use the power rule
thanks for your help, i understand how to get 12x^3 but getting a bit lost on what happens to -3√x+8 ??????
Originally Posted by bart 1000 thanks for your help, i understand how to get 12x^3 but getting a bit lost on what happens to -3√x+8 ?????? √x = x^0.5 So, lower the power by 1 to make it -1/2, then multiply the co-efficient by 1/2 to make it -1.5
Originally Posted by bart 1000 thanks for your help, i understand how to get 12x^3 but getting a bit lost on what happens to -3√x+8 ?????? Do you mean? $\displaystyle \sqrt{x+8}=(x+8)^{\frac{1}{2}}$ If so $\displaystyle \frac{d}{dx}(x+8)^{\frac{1}{2}}=\frac{1}{2}(x+8)^{-\frac{1}{2}}=\frac{1}{2\sqrt{x+8}}$
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