Question

If f(x)=2g(x)+h(x)+k, where k is constant, then
Select one:
a. f'(x)=g(x)
b. f'(x)=g'(x)+k
c. f'(x)=2g'(x)+h(x)
d. f(x)=0​

Answer 1

Topic :-

Differentiation

To Differentiate :-

f(x) = 2g(x) + h(x) + k

where k is constant

Concept Used :-

Differentiation of a given function f(x) is given as f'(x), means function gets derivated with respect to 'x'.

For example,

f(x) = x² + 2x + 3 then

$\tt {\dfrac{d(f(x))}{dx}=\dfrac{d(x^2)}{dx}+\dfrac{d(2x)}{dx}+\dfrac{d(3)}{dx}}$

$\tt {f'(x) = 2x + 2}$

Differentiation of a constant is '0'.

Solution :-

f(x) = 2g(x) + h(x) + k

Differentiating both sides,

$\tt {\dfrac{d(f(x))}{dx}=\dfrac{d(2g(x))}{dx}+\dfrac{d(h(x))}{dx}+\dfrac{d(k)}{dx}}$

$\tt {\dfrac{d(f(x))}{dx}=2\dfrac{d(g(x))}{dx}+\dfrac{d(h(x))}{dx}+\dfrac{d(k)}{dx}}$

$\tt{f'(x)=2g'(x)+h'(x)+0}$

$\tt{f'(x)=2g'(x)+h'(x)}$

Answer :-

So, answer is f'(x) = 2g'(x) + h'(x)

Additional Formulae :-

$\tt {\dfrac{d(sinx)}{dx}= cosx}$

$\tt {\dfrac{d(cosx)}{dx}= -sinx}$

$\tt {\dfrac{d(tanx)}{dx}= sec^2x}$

$\tt {\dfrac{d(cosecx)}{dx}= -cosecx.cotx}$

$\tt {\dfrac{d(secx)}{dx}= secx.tanx}$

$\tt {\dfrac{d(cotx)}{dx}= -cosec^2x}$