find the equation of tangent to the curve y=2(x^1/2) at (1,2)

since i am new to matlab plese help me out with this assignment

find the equation of tangent to the curve y=2(x^1/2) at (1,2)

[KSSV]

You can do this with ease.....you need to do hand calculation mostly. What have you attempted?

[Nishanth G]

syms s(t) f(x)
f(x)=input("enter the function")
m=diff(f,x)
x1=input("enter the value of x for slope point")
y1=input("enter the value of y for slope point")


y=m*(x-x1)+y1
print(y

i got till this i need to apply x1 value in m , how do i do that ?

[Nishanth G]

thanks got it

Answer 1

[Stephan]

https://de.mathworks.com/help/symbolic/syms.html#buoeaym-1

https://de.mathworks.com/help/symbolic/diff.html

[Stephan]

syms f(x)
f(x) = 2*(x^(1/2)) % symbolic function
Df = diff(f,x) % derivative
m = Df([1 2]) % values of slope for x=1 & x=2 in a vectorized manner
m_vals = double(m) % convert symbolic (exact) results to numeric