Question

Compute for the derivative of the function below for x = 2. h(x)=(log2x2)(32x3)

Answer 1

Answer:

h'(2) = 256 log₂e + 768

Step-by-step explanation:

Hi,

Given h(x) = log₂x². (32x³)

Using product rule to find derivatives

If h(x) = f(x).g(x), then h'(x) = f'(x)g(x) + f(x)g'(x)

So, if we choose f(x) = log₂x² and g(x) = 32x³, then

h'(2) = f(2)g'(2) + f'(2)g(2)---------------(*)

Let f(x) = log₂x² = 2log₂x = 2 loge(x)/loge(2) = (2/loge(2))loge(x)

f'(x) = 2/xloge(2)

f(2) = log₂2² = 2

f'(2) = log₂e

Let g(x) = 32x³

g'(x) = 96x²

g(2) = 256

g'(2) = 384,

On substituting the above values in (*), we get

h'(2) = 256 log₂e + 768

Hope, it helped !

Answer 2

Solution:

$\frac{d {x}^{n} }{dx} = n {x}^{n - 1} \\ \\ \frac{d log_{2}(x) }{dx} = \frac{1}{ ln(2)x }$
So

$h(x) = log_{2}( {x}^{2} ) (32 {x}^{3} ) \\ \\ {h}^{'} (x) = log_{2}( {x}^{2} )(96 {x}^{2} ) + (32 {x}^{3} ) \frac{2}{ ln(2)x }$

So to find derivative at x= 2 put the value in derivative
${h}^{'} (2) = log_{2}( {2}^{2} )(96 ({2})^{2} ) + (32 ({2})^{3} ) \frac{2}{ 2ln(2) } \\ \\ = 2 \times 4 \times 96 + \frac{32 \times 8}{ ln(2) } \\ \\ {h}^{'} (2)= 768 + \frac{256}{ ln(2) }$

Hope it helps you.