If 48+8f(x)+x2(f(x))3=0 and f(2)=−2, find f′(2).

Answer 1

Hi,

48+8f(x)+x2(f(x))3=0 and f(2)=−2, find f′(2).

1) take a derivative of 48+8f(x)+x2(f(x))3=0

0 +8f'(x) + 2x[f(x)]^3 + x^2[3 f(x)^2 f'(x) ] = 0

Now replace x = 2 0 +8f'(2) + 2(2) [f(2)^3 ]+ 2^2[ 3 f(2)^2 f'(2)] = 0

0 +8f'(2) - 32 + 4[ 12f'(2)] = 0 now solve for f'(2)

56 f'(2) = 32

f'(2) = 4/7

I hope it is useful,

Minoo

Answer 2

Assuming the problem is:

Take the derivative of each element in the equation.

The derivative of 48 is zero

The derivative of 8•f(x) is 8•f '(x)

To take the derivative of x²(f(x))³, since this is the product of two items (x² and (f(x))³) you must use the product rule which is (u•v)' = u'v + uv'. Let u = x² and let v = (f(x))³. Then u' (using the power rule) is 2x and v' (using both the power rule and the chain rule) is 3(f(x))²•f '(x). Putting these together, the derivative of x²(f(x))³ is (2x)(f(x))³ + (x²)(3(f(x))²•f '(x)

The derivative of zero is zero.

Putting these all together we get:

0 + 8•f '(x) + (2x)(f(x))³ + (x²)(3(f(x))²•f '(x) = 0

8•f '(x) + (2x)(f(x))³ + (3x²)(f(x))²•f '(x) = 0

8•f '(x) + (3x²)(f(x))²•f '(x) = -(2x)(f(x))³

f '(x)[8 + (3x²)(f(x))²] = -(2x)(f(x))³

f '(x) = -(2x)(f(x))³/[8 + (3x²)(f(x))²]

Now, plug in x = 2 and f(2) = -2 to find the answer

Answer 3

Answer: f'(2) = 6/5

Explanation:

We have 48+8f(x)+3x²f(x) = 0.

Next, we apply implicit differentiation with respect to x:

8f'(x) + 6xf(x) + 3x²f'(x) = 0.

Next, we let x=2, to get:

8f'(2) + 6(2)f(2) + 3(2)²f'(2) = 0.

Next, we substitute f(2) = -2 and simplify:

8f'(2) + 12(-2) + 12f'(2) = 0, so

20f'(2) = 24.

Finally, we solve for f'(2),

f'(2) = 24/20 = 6/5.