How do you differentiate $f(x) = sqrt (3^x) / x^2$ ?

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Answer 1 · 2017-06-08T14:24:03

$f'(x)={sqrt(3^x)(xln3-4)}/(2x^3), or,$

$f'(x)={(sqrt(3^x))(ln(3^x)-4)}/(2x^3).$

$f(x)=sqrt(3^x)/x^2=(3^x)^(1/2)/x^2=(3^(1/2))^x*x^-2=(sqrt3)^x*x^-2.$

Now, we know that,

$(1): F(x)=G(x)H(x) rArr F'(x)=G(x)H'(x)+G'(x)H(x).$

$(2): (a^x)'=a^xlna.$

$(3) : (x^n)'=nx^(n-1).$

$:. f'(x)=(sqrt3)^x(x^-2)'+(x^-2)((sqrt3)^x)'.$

$=((sqrt3)^x(-2x^(-2-1))+(x^-2){(sqrt3)^xlnsqrt3}$

$=-2x^-3sqrt(3^x)+{sqrt(3^x)ln3}/(2x^2)$

$=-{2sqrt(3^x)}/x^3+{sqrt(3^x)ln3}/(2x^2)$

$rArr f'(x)={sqrt(3^x)(xln3-4)}/(2x^3), or,$

$f'(x)={(sqrt(3^x))(ln(3^x)-4)}/(2x^3).$

Enjoy Maths.!

How do you differentiate f(x) = sqrt (3^x) / x^2 f(x) = sqrt (3^x) / x^2?