Q) Given that f(x) =(5x -4)½ , evaluate f ' (4)
Can I get a step by step proccess please?
+33654 f(x)=(5x-4)^\frac{1}{2}f'(x)=\frac{1}{2}(5x-4)^{-\frac{1}{2}}*5=\frac{5}{2(5x-4)^{\frac{1}{2}}}f'(4)=\frac{5}{2(5*4-4)^{\frac{1}{2}}}=\frac{5}{2(20-4)^\frac{1}{2}}=\frac{5}{2*\sqrt{16}}=\frac{5}{2*4}=\frac{5}{8}$$
.
Thanks so much! However for the first derrivative why are you "*5" I do not understand. Where are you getting it from?
+33654 Ok. If I have f = (a*x+b)c and want to differentiate say with respect to x, I can first set z = a*x+b and I have f = zc. I can differentiate this with respect to z to get df/dz = c*zc-1. But I want it differentiated with respect to x not z, so I must use the chain rule, which says that df/dx = df/dz*dz/dx.
Now, we have found df/dz, so we need dz/dx. Since z = a*x+b then dz/dx = a. Hence overall we have df/dx = df/dz*dz/dx = c*zc-1*a = c*(a*x+b)c-1*a, where I've replaced z by its definition in terms of x.
In your case we have a = 5; b = -4; and c = 1/2, so in your case df/dx = (1/2)*(5*x-4)-1/2*5.
I did this all in one step. I'm so used to doing this I don't need to break it down into separate steps, but clearly I should have done here!
If this still isn't clear, let me know.
This is a big help, thanks. However, I am still having difficulty understanding as to how and why I can get the "z" equation, for instance what happens to "c"?
+33654 c is a constant which is just 1/2 in your case.
However, let me do it with just your numbers:
f = (5x - 4)1/2
Let z = 5x - 4 ...(1) so
f = z1/2
Differentiate f with respect to z
df/dz = (1/2)*z(1/2)-1 or df/dz = (1/2)*z-1/2 ...(2)
We want df/dx not df/dz, so we use the chain rule df/dx = df/dz*dz/dx ...(3)
We have df/dz, so we need dz/dx
From (1) above we find dz/dx = 5 ...(4)
Put (2) and (4) into (3) to get
df/dx = df/dz*dz/dx = (1/2)*z-1/2*5 = 5/(2*z1/2)
Replace z using (1) to get
df/dx = 5/(2*(5x-4)1/2)
Is this any clearer?
