How do you find the Slope of the curve y=sqrt(x) at the point where x=4?

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The swell of the incurvation ##y=f(x)## at ##x=a## is ##f'(a)##. Let us appear at the posted drift.

By rewriting the square-root as 1/2-power, ##f(x)=sqrt{x}=x^{1/2}##

By the effectiveness government, ##f'(x)=frac{1}{2}x^{ -1/2}=frac{1}{2sqrt{x}}##

By plugging in ##x=4##, Slope: ##m=f'(4)=frac{1}{2sqrt{4}}=frac{1}{4}##

Hence, the swell is ##1/4##.

I anticipation that this helps.

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