Find the derivative of h(x)=7lnx^4

h'(x)=???

is it 24x^3/x^4??

I mean 28x^3/x^4

let z = x^4 so dz/dx = 4 x^3

then
y = 7 ln z
dy/dz = 7/z
dy/dx = dy/dz dx/dx = (7/z)(4 x^3)
= 28 (x^3/x^4)
= 28/x

Calculus I mean.

you could also notice that

ln(x^4) = 4 lnx
so,

y = 7*4 lnx = 28 lnx
dy/dx = 28/x

To find the derivative of h(x) = 7ln(x^4), we can use the chain rule, which states that if we have a function of the form f(g(x)), then its derivative is given by f'(g(x)) * g'(x).

In this case, f(u) = 7ln(u) and g(x) = x^4. To find the derivative of h(x), we need to find the derivative of the inner function, g(x), and then multiply it by the derivative of the outer function, f'(u).

The derivative of g(x) = x^4 is obtained by applying the power rule, which states that the derivative of x^n is n * x^(n-1). In this case, n = 4, so the derivative of g(x) is:

g'(x) = 4 * x^(4-1) = 4x^3

The derivative of f(u) = 7ln(u) can be found using the logarithmic differentiation. The derivative of ln(u) with respect to u is 1/u, so applying the chain rule, the derivative of f(u) = 7ln(u) is:

f'(u) = 7 * (1/u)

Now, we can substitute back u = x^4, and we get:

f'(u) = 7 * (1/x^4)

Finally, we can multiply the derivatives of the inner and outer functions together to find the derivative of h(x):

h'(x) = f'(u) * g'(x) = 7 * (1/x^4) * 4x^3

Simplifying, we obtain:

h'(x) = 28x^3 / x^4 = 28/x