Find the slope of the tangent line to the curve y = 2(ln x)^x/2 at x = e. Find t

Question

Find the slope of the tangent line to the curve y = 2(ln x)^x/2 at x = e. Find the equation of the tangent line in slope-intercept form to the curve x + squareroot xy = 6 at (4, 1). If two resistors of R_1 and R_2 ohms are connected in parallel in an electric circuit to make an R-ohm resistor, the value of R can be found from the equation 1/R = 1/R_1 + 1/R_2. If R_2 is decreasing at the rate of 1 hm/sec and R_2 is increasing at the rate of 0.5 ohm sec. at what rate is R changing when R_1 = 75 ohms and R_1 = 50 ohms?

Explanation / Answer

4)given curve y=2(lnx)x/2

apply logaritham on both sides

lny=ln[2(lnx)x/2]

lny =ln2 +ln(lnx)x/2

lny =ln2 +(x/2)ln(lnx)

differentiate with respect to x on both sides

(1/y)(dy/dx)=0 +(1/2)ln(lnx) +(x/2)(1/(lnx))(1/x)

(1/y)(dy/dx)= (1/2)ln(lnx) +(1/2)(1/(lnx))

(1/y)(dy/dx)= (1/2)[ln(lnx) +(1/(lnx))]

dy/dx= y(1/2)[ln(lnx) +(1/(lnx))]

dy/dx=2(lnx)x/2(1/2)[ln(lnx) +(1/(lnx))]

dy/dx=(lnx)x/2[ln(lnx) +(1/(lnx))]

at x =e

dy/dx=(lne)e/2[ln(lne) +(1/(lne))]

dy/dx=(1)e/2[ln(1) +(1/1)]

dy/dx =1[0 +1]

dy/dx =1

slope of the tangent at x =e is 1

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