Question
Question: How do you differentiate \(y=4\ln x+2\cos x-3{{e}^{x}}\)?...
How do you differentiate y=4lnx+2cosx−3ex?
Solution
Now to differentiate the given function we will use the property of differentiation which states dxd(f±g)=dxd(f)±dxd(g) . Now we will use the property dxd(cf)=cdxdf and take constant out of all the terms. Now we know that the differentiation of cosx is given by −sinx , the differentiation of ex is given by exand the differentiation of lnx=x1 . Hence using this we get the differentiation of the given function.
Complete step-by-step answer:
Now consider the given function y=4lnx+2cosx−3ex .
To differentiate the function we will find the value of dxdy .
Now to differentiate the given function let us first understand some properties of differentiation.
Now we know that the differentiation of addition or subtraction of two function is given by ⇒dxd(f+g)=dxd(f)+dxd(g) and dxd(f−g)=dxd(f)−dxd(g)
Hence using this we can write the given differential as
⇒dxdy=dxd(4lnx)+dxd(2cosx)−dxd(3ex)
Now we know that for differentiation of any function f dxd(cf)=cdxdf where c is any constant.
Hence in the above equation we will take out the respective constants in each term.
⇒dxdy=4dxd(lnx)+2dxd(cosx)−3dxd(ex)
Now we have the differentiation of cosx is given by −sinx, the differentiation of lnx is x1 and the differentiation of ex is given by ex .
Hence using this values of differentiation we get,
⇒dxdy=x4−2sinx−3ex
Hence the differentiation of the given function is x4−2sinx−3ex
Note: Now note that the differentiation we have the differentiation of the sum of functions is sum of differentiation of function. But note that this is not the same case with multiplication and division. For multiplication we have dxd(f.g)=gdxdf+fdxdg . Similarly of we have for function in division we have the differentiation of the function as (gf)′=g2f′g−g′f . Also we can use multiplication rule to find the differentiation of fractions by considering dxdf.dxd(g1) .