Logarithmic Differentiation?! HELP?

Question

We have the function f(x) = (2 x^2 - x +10)^sin(x). Use logarithmic differentiation to find f'(x).

f('1) =

I keep getting 10.32 as an answer, but apparently, it's wrong.

Answer 1

Say y=(2x^2-x+10)^sinx then
lny=sinx*ln(2x^2-x+10)
takin derivatives of both sides
y'/y={cosx*ln(2x^2-x+10)+
sinx*[(4x-1)/(2x^2-x+10)]}
So y'=y *{.......................}

Answer 2

regularly, logarithmic differentiation may well be used effectively if the logarithm of the function could be simplified utilising the regulations of logarithms. one way of writing logarithmic differentiation is during the formulation f'(x)=f(x)*d/dx[ ln f(x) ]. working example, if f(x)=x^sin x, then ln f(x)=(sin x)(ln x) is less complicated than f(x) is. Then the by-made from ln f(x) is (sin x)(one million/x)+(cos x)(ln x), so f'(x)=x^(sin x) *[(sin x)/x +(cos x)(ln x)].

Answer 3

take the ln of both sides
ln f(x)=ln[(2x^2-x+10)^sinx]
However, properties of logs said that we can bring down exponents as coefficients.
ln f(x)=sinx*ln(2x^2-x+10)
take the deriv of both sides
f'(x)/f(x)= cosx*ln(2x^2-x+10)+(4x-1)(sinx)/(2x^2-x+10)
multiply both sides by f(x)
f'(x)= [(2 x^2 - x +10)^sin(x)][cosx*ln(2x^2-x+10)+(4x-1)(sinx)/(2x^2-x+10)]





f('1) =11.47

Answer 4

I got 16.7759

It's a question of taking logs on both sides:

log (y) = log RHS

Then differentiation

1/y dy/dx = differential of RHS (use chain rule)

Then multiply both sides by y (your original function)

dy/dx = diff. of RHS multiplied by orig. function.

Make sure you're using radians and not degrees when you plug in your trig functions

It's a complex mess, and I hope my answer is right.

Answer 5

ln f = sin(x)*ln(2x^2 - x + 10)

take derivative

1/f*df/dx = sin(x)*[(4x - 1)/(2x^2 - x + 10)] + cos(x)*ln(2x^2 - x + 10)

so df/dx = f*[sin(x)*[(4x - 1)/(2x^2 - x + 10)] + cos(x)*ln(2x^2 - x + 10)]

let g = [sin(x)*[(4x - 1)/(2x^2 - x + 10)] + cos(x)*ln(2x^2 - x + 10)]

df/dx = f*g
f(1) = 7.5215
g(1) = 1.5251
df/dx = 11.4708

Answer 6

ln f = sinx* ln(2x^2-x+10)
f´/f= sin(x) *(4x-1)/(2x^2-x+10) +cos(x)ln (2x^2-x+10)
so
f´= (2x^-x+10) ^sin(x)* [(4x-1)/(2x^2-x+10)*sinx +ln(2x^2-x+10)*cos x]