For problems 4-8, find the derivative. Simplify where possible. y = log_5 (4 - x

Question

For problems 4-8, find the derivative. Simplify where possible. y = log_5 (4 - x^2) y = sin (tan square root 1 + x^3) y = e^-5x cos 3x y = sin^-1 (2x + 1) g(x) = In(itanh 2x) A trough is 20 feet long and its ends have the shape of an isosceles triangle that are 8 feet across at the top and have a height of 10 feet. If the trough is being filled with water at the rate of 4 ft^3 / min, how fast is the water level rising when the water level is 1 ft deep? Find a formula for the inverse of y = ln(x + 5). Find the limit. State the indeterminate type where necessary before applying L'Hospital's Rule. lim_x rightarrow infinity ln x / sin (pi x) lim_x rightarrow 0+ sin x ln x lim_x rightarrow 0 (csc x - cot x) lim_x rightarrow 0+ (cos x)^1/x^2

Explanation / Answer

4)y =log5(4-x2)

y =(ln(4-x2))/(ln5)

y =(1/ln5)(ln(4-x2))

derivative of lnp= 1/p ,derivative of f(g(x))= f '(g(x)) *g'(x)

dy/dx=(1/ln5)(1/(4-x2))*(0-2x)

dy/dx=(1/ln5)(-2x/(4-x2))

5)y =sin(tan((1+x3)1/3)

derivative of sinp =cosp, derivative of tanp=sec2p , derivative of xn=nxn-1,derivative of f(g(x))= f '(g(x)) *g'(x)

dy/dx=(cos(tan((1+x3)1/3)) *(sec2((1+x3)1/3))*((1/3)((1+x3)-2/3))(0+3x2)

dy/dx=(cos(tan((1+x3)1/3)) *(sec2((1+x3)1/3))*(x2(1+x3)-2/3)

dy/dx=(x2(1+x3)-2/3)*(sec2((1+x3)1/3))(cos(tan((1+x3)1/3))

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