Answer:
(a) k'(0) = f'(0)g(0) + f(0)g'(0)
(b) m'(5) = 
Step-by-step explanation:
(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., ![\frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Cmathrm%7Bd%7D%20%5Bf%28x%29%5Ctimes%20g%28x%29%5D%7D%7B%5Cmathrm%7Bd%7D%20x%7D%3D%5Cfrac%7B%5Cmathrm%7Bd%7D%20f%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D%5Ctimes%20g%28x%29%20%2B%20f%28x%29%5Ctimes%5Cfrac%7B%5Cmathrm%7Bd%7D%20g%28x%29%7D%7B%5Cmathrm%7Bd%7D%20x%7D)
this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>
on substituting the value x=0, we will get the value of k'(0)
{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}
(b) m(x) is a function of two functions f(x) and g(x) [
]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : 
this will give <em>
</em>
on substituting the value x=5, we will get the value of m'(5).
{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}
{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }
{ NOTE :
}
The answer for that problem is z =12+6
Answer:
Triangle APB is an isosceles triangle ⇒ 3rd answer
Step-by-step explanation:
* Lets explain the how to solve the problem
- ABCD is a square
∴ AB = BC = CD = AD
∴ m∠A = m∠∠B = m∠C = m∠D = 90°
- DPC is equilateral triangle
∴ DP = PC = DC
∴ m∠DPC = m∠PCD = m∠CDP = 60°
- In the Δs APD , BPC
∵ AD = BC ⇒ sides of the square
∵ PD = PC ⇒ sides of equilateral triangle
∵ m∠ADB = m∠BCP = 30° (90° - 60° = 30) ⇒ including angles
∴ Δs APD , BPC are congregant ⇒ SAS
- From congruent
∴ AP = BP
∴ Triangle APB is an isosceles triangle