Differentiate x^(1/2)ln(3x) with respect to x.
First we notice that this is a product of two functions of x, so we are going to use the product rule. Recall (uv)'(x)=u'(x)v(x)+v'(x)u(x). Let u(x)=x^(1/2) and v(x)=ln(3x). We need to find u'(x) and v'(x). We have that u'(x)=(1/2)x^(-1/2) by simple differentiation. Also v'(x)=3/3x=1/x by applying the chain rule. Therefore (uv)'(x)=(1/2)x^(-1/2)*ln(3x)+(1/x)*x^(1/2)=(1/2)x^(-1/2)ln(3x)+x^(-1/2)=x^(-1/2)((1/2)ln(3x)+1), simplifying it down to its simplest form.
AR
Related Maths A Level answers
All answers ▸express (3x + 5)/(x^2 + 2x - 15) - 2/(x - 3) as a single fraction its simplest form
Find the finite area enclosed between the curves y=x^2-5x+6 and y=4-x^2
f(x)=2x^3-7x^2+4x+4, prove that (x-2) is a factor and factorise f(x) completely
Express cos(x) + (1/2)sin(x) in terms of a single resultant sinusoidal wave of the form Rsin(x+a)